Answers
In order for the lever to balance, F1D1 must be equal to F2D2.
To determine what must be equal to F2D2 in order for the lever to balance, we need to understand the principle of a lever and how it works.
A lever is a simple machine consisting of a rigid beam (in this case, represented by the drawing) that pivots around a fulcrum. The lever operates on the principle of torque, which is the rotational force produced when a force is applied at a distance from the fulcrum.
In the drawing, there are two forces acting on the lever: F1 and F2. F1 is applied at a distance D1 from the fulcrum, while F2 is applied at a distance D2 from the fulcrum. To balance the lever, the clockwise torque produced by F1 must be equal to the counterclockwise torque produced by F2.
The torque produced by a force is calculated by multiplying the force by the distance from the fulcrum. Mathematically, it can be represented as:
Torque = Force × Distance
For the lever to balance, the torques on both sides must be equal. Therefore, we have the equation:
F1 × D1 = F2 × D2
In other words, F2D2 must be equal to F1D1 for the lever to balance.
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Related Questions
The force of Earths gravity keeps earth in orbit true or false
Answers
Answer:
True
Explanation:
The force of gravity keeps all of the planets in orbit around the sun
True. The force of gravity keeps all of the planets in orbit around the sun.
What is Gravity?
The force that pulls items toward the center of a planet or other entity is called gravity. All of the planets are kept in orbit around the sun by gravity.
Gravity applies to everything that has mass. Gravity is stronger for objects with higher mass. Along with distance, gravity weakens as well. Therefore, the gravitational pull of two things becomes stronger the closer they are to one another.
The mass of the Earth is what creates gravity. The combined gravitational force of all of its mass acts on the mass in your body.
Therefore, True. The force of gravity keeps all of the planets in orbit around the sun.
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Calculate the absolute pressure at an ocean depth of 1.0 x 10³ m. Assume that the density of the water is 1.025 x 10³ kg/m³ and that Po = 1.01 x 10^5 Pa.
Answers
The absolute pressure at an ocean depth of 1.0 x 10^3 m is 1.002 x 10^8 Pa.
What is hydrostatic pressure?Hydrostatic pressure is the pressure that a fluid exerts on a surface due to the weight of the fluid above it. It is the result of the force of gravity acting on a column of fluid, and it is directly proportional to the height of the column of fluid and the density of the fluid.
The absolute pressure at an ocean depth of 1.0 x 10^3 m can be calculated using the hydrostatic pressure equation:
P = ρgh + Po
where:
P is the absolute pressure at the given depth
ρ is the density of the water
g is the acceleration due to gravity (assumed to be 9.81 m/s²)
h is the depth of the ocean
Po is the atmospheric pressure at the surface (assumed to be 1.01 x 10^5 Pa)
Substituting the given values, we get:
P = (1.025 x 10^3 kg/m³) x (9.81 m/s²) x (1.0 x 10^3 m) + 1.01 x 10^5 Pa
P = 1.025 x 9.81 x 10^6 Pa + 1.01 x 10^5 Pa
P = 1.002 x 10^8 Pa.
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A 0.530-kg cart moving at 0.572 m/s to the right collides elastically with a 0.25-kg cart initially at rest. The 0.25-kg cart then moves off rapidly and compresses a spring before the 0.530-kg cart can catch it again.
Answers
To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy.First, let's find the velocity of the 0.530-kg cart after the collision. We can use the conservation of momentum:m1v1 + m2v2 = m1v1' + m2v2'where m1 and v1 are the mass and velocity of the 0.530-kg cart before the collision, m2 and v2 are the mass and velocity of the 0.25-kg cart before the collision, and v1' and v2' are the velocities of the carts after the collision.Plugging in the numbers, we get:(0.530 kg)(0.572 m/s) + (0.25 kg)(0 m/s) = (0.530 kg)v1' + (0.25 kg)v2'Solving for v1', we get:v1' = [(0.530 kg)(0.572 m/s) + (0.25 kg)(0 m/s)] / (0.530 kg + 0.25 kg) = 0.378 m/s to the rightSo the 0.530-kg cart moves off to the right at 0.378 m/s after the collision.Next, let's find the maximum compression of the spring. We can use the conservation of kinetic energy:(1/2)m2v2^2 = (1/2)kx^2where k is the spring constant and x is the maximum compression of the spring.We know the mass and velocity of the 0.25-kg cart before the collision (v2 = 0 m/s), so we can solve for k:k = 2(1/2)m2v2^2 / x^2 = m2v2^2 / x^2Plugging in the numbers, we get:k = (0.25 kg)(0 m/s)^2 / x^2 = 0This means that the spring constant is 0, which is not physically possible. Therefore, there must be an error in the problem statement or some missing information that would allow us to calculate the maximum compression of the spring.
A block of wood is attached to a very lightweight metal rod, which is attached to a fixed pivot point on a table. The block is able to slide on the table with negligible friction, and the pivot is also free to rotate with negligible friction. The block's mass is M and the rod's length is ℓ. A bullet is moving parallel to the table and perpendicular to the rod when it collides and embeds within the block. The bullet's speed just before entering the block is v and its mass is m.
1. Find the angular momentum of the combined bullet–block system about the vertical pivot axis. (Use any variable or symbol stated above as necessary. Enter the magnitude.)
2. Find the fraction of the original kinetic energy of the bullet that is converted into internal energy within the bullet-block system during the collision. (Use any variable or symbol stated above as necessary.)
Answers
1. The angular momentum of the combined bullet-block system about the vertical pivot axis is 0.
2. The fraction of the original kinetic energy of the bullet converted into internal energy within the bullet-block system during the collision is given by [m * v² - (M + m) * V²] / [m * v²].
1. To find the angular momentum of the combined bullet-block system about the vertical pivot axis, we need to consider the initial and final angular momentum.
Initially, before the collision, the bullet has no angular momentum about the pivot axis since it is moving parallel to the table and perpendicular to the rod.
After the collision, when the bullet embeds within the block, the combined bullet-block system starts rotating about the pivot axis due to the conservation of angular momentum.
The angular momentum of the system can be calculated using the formula:
Angular momentum = moment of inertia × angular velocity
The moment of inertia of the system depends on the distribution of mass and the axis of rotation. Assuming the block and bullet have negligible rotational inertia compared to the rod, we can consider the moment of inertia to be that of the rod.
The moment of inertia of a rod rotating about one end (pivot) is given by:
I = (1/3) * M * ℓ²
where M is the mass of the block, and ℓ is the length of the rod.
The angular velocity (ω) can be determined by considering the conservation of angular momentum:
Initial angular momentum = Final angular momentum
0 = (1/3) * M * ℓ² * ω
Since the initial angular momentum is zero, the final angular momentum of the system is also zero.
Therefore, the angular momentum of the combined bullet-block system about the vertical pivot axis is 0.
2. To find the fraction of the original kinetic energy of the bullet that is converted into internal energy within the bullet-block system during the collision, we can use the principle of conservation of kinetic energy.
The initial kinetic energy of the bullet before the collision is given by:
Initial kinetic energy = (1/2) * m * v²
After the collision, the bullet embeds within the block, and both the bullet and the block gain internal kinetic energy due to their rotational motion.
The final kinetic energy of the bullet-block system is given by:
Final kinetic energy = (1/2) * (M + m) * V²
where V is the final velocity of the combined bullet-block system after the collision.
Since the bullet and block are now rotating about the pivot axis, part of the initial kinetic energy is converted into internal rotational kinetic energy.
The fraction of the original kinetic energy converted into internal energy can be calculated as:
Fraction of kinetic energy converted = (Initial kinetic energy - Final kinetic energy) / Initial kinetic energy
Substituting the values:
Fraction of kinetic energy converted = [(1/2) * m * v² - (1/2) * (M + m) * V²] / [(1/2) * m * v²]
Simplifying the equation, we can cancel out common terms:
Fraction of kinetic energy converted = [m * v² - (M + m) * V²] / [m * v²]
Therefore, the fraction of the original kinetic energy of the bullet converted into internal energy within the bullet-block system during the collision is given by [m * v² - (M + m) * V²] / [m * v²].
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Two glasses are filled with liquid to the same height . Glass W is filled with water and glass F is filled with fruit juice. Fruit juice has a higher density than water. Will the pressure in the glass of water be higher or smaller than the pressure in the glass of fruit juice.? Explain your answer
Answers
The weight at the foot of each glass will be the same because it is decided by the stature of the fluid column over it and the increasing speed due to gravity. Be that as it may, the weight applied by the fluid on the dividers of the holder will be diverse due to contrasts in thickness.
Since natural product juice incorporates a higher thickness than water, the mass of the same volume of natural product juice will be more noteworthy than that of water.
This implies that the constraint applied by the natural product juice on the dividers of the glass will be more prominent than the constraint applied by the water on the dividers of its glass, as the drive is straightforwardly corresponding to mass.
Subsequently, the weight of the glass of natural product juice will be higher than the weight of the glass of water.
In rundown, in spite of the fact that the weight at the foot of each glass will be the same, the weight applied by the fluid on the dividers of the holder will be diverse due to the contrasts in thickness, with the natural product juice applying a more noteworthy weight.
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STEM
Science Challenge
Rockin’ Around the Park
You are a geologist who teaches at a university. You receive the following letter from a local park
ranger.
Dear Professor Rubble,
Our park has some magnificent rock formations along our hiking trails. We receive many questions
from park visitors about how these rock formations were created, and how long it took for them to
form. One visitor even claimed that the same exact rock formation existed near her home thousands
of kilometers away. How can this be possible?
To answer these questions, can you help us develop a model of the Earth processes that shape and
change Earth’s surface that we can display in our nature center? We would also like an article to
accompany the model. The article should explain the model, as well as the time and spatial scales
involved in the processes that form and change rocks on Earth’s surface.
Yours sincerely,
Flint Cobblestone
Q) 1 :How is it possible that identical rock formations can exist thousands of kilometers apart?
Q) 2 :What clues would you look for in the identical rock formations to prove that they formed at the same
time? How might you incorporate this information in your model?
Q) 3 :What processes could have separated the rock formations over time?
Q) 4 :How might you incorporate this information in your model?
Q) 5 :The park contains mountain ranges. How could you determine how and when the mountain ranges
formed?
Q) 6 : How could you incorporate this information in your model?
Q) 7 : Do weathering and erosion cause fast or slow changes to Earth’s surface? Do they affect large or
small areas?
Some types of weathering and erosion occur quickly. A landslide, for example, can quickly transport
and break down Earth materials. Other types of weathering and erosion take place over millions of
years. These processes occur on both large scales (example, mountain ranges) and small scales
(example: a boulder). You could have two separate models to show different time and spatial scales.
One model might show the slow weathering and erosion of a rock using drops of water to simulate
rainfall. The second model might show the fast weathering and erosion of a mountainside during a
landslide triggered by an earthquake.
Q) 8 : How might you include this information in your model?
To model weathering, you could put rocks in a container and shake it many times. You could use
water or a fan to model erosion by rivers or wind.
Q) 9 : Which rock cycle processes will you include in your model? How will you model the sources of energy
that drive these processes?
Models should include weathering, erosion, deposition, compaction, cementation, melting, cooling,
heat, and pressure. Students might use a fan to model wind energy or a source of heat to model
energy from Earth’s interior.
Q) 10 : How will you model processes that occur on different time and spatial scales?
Develop Your Model
Use your information from the questions to construct your model and
prepare your article for the park to display at their nature center.
Models will vary, but should include processes that describe the cycling of Earth materials and the
flow of energy that drives the cycling. Examples of processes include weathering, erosion, deposition,
compaction, cementation, melting, crystallization, pressure, deformation, subduction, and seafloor
spreading.
Evaluate Your Model
After completing your model, identify the model elements in the table below.
Model Elements Descriptions
Components
(What are the different
parts of my model?)
Relationships
(How do the components
of my model interact?)
Connections
(How does my model help
me understand the
phenomenon?)
As the final part of your homework, construct an article that will accompany your model at the park’s
nature center. Your article should explain how Earth processes shape and change rocks on Earth’s
surface at different time and spatial scales.
Outline your article in the space below. Organize your central ideas into a structure that facilitates
understanding. Next, add relevant content including the evidence and reasoning that supports each
central idea. When you are finished with your outline, write your article in your Science Notebook.
Students’ articles should show an understanding that geoscience processes have changed Earth’s
surface at varying time and spatial scales. They should use evidence and reasoning to construct their
explanations for how the mountains and the rock formations in the park formed. They should identify
and describe the time and spatial scales involved (slow, large-scale, fast, small-scale).
Congratulations! You have reached the end of the science challenge.
Answers
Answer:
Hope this helps ;) don't forget to rate this answer !
Explanation:
It is possible that identical rock formations can exist thousands of kilometers apart because of the rock cycle, a process that involves the continuous transformation of rocks through various stages such as weathering, erosion, deposition, and more.
To prove that the identical rock formations formed at the same time, geologists can look for clues such as the presence of the same type of minerals, the same layering or structure, and similar levels of weathering or erosion. This information can be incorporated into the model by including representations of these clues and explaining their significance in the rock cycle.
Processes that could have separated the rock formations over time include tectonic movement, erosion, and weathering. These processes can be incorporated into the model by including representations of tectonic plates and showing how they can move and collide, as well as by including examples of erosion and weathering and explaining their role in the rock cycle.
To determine how and when mountain ranges formed, geologists can study the rock formations, the types of minerals present, and the levels of weathering and erosion. This information can be incorporated into the model by including representations of different types of rock formations and explaining how they were formed through processes such as mountain building and erosion.
Weathering and erosion can cause both fast and slow changes to Earth's surface, and can affect both large and small areas. To include this information in the model, you could have two separate models to show different time and spatial scales. One model might show the slow weathering and erosion of a rock, while the second model might show the fast weathering and erosion of a mountainside during a landslide.
To model weathering and erosion, you could put rocks in a container and shake it many times to simulate weathering, or use water or a fan to model erosion by rivers or wind. You could also use a source of heat to model energy from Earth's interior, or a fan to model wind energy.
In your model, you should include processes that describe the cycling of Earth materials and the flow of energy that drives the cycling. These processes include weathering, erosion, deposition, compaction, cementation, melting, crystallization, pressure, deformation, subduction, and seafloor spreading.
In your article, you could start by introducing the rock cycle and explaining the various processes involved. You could then describe how these processes shape and change rocks on Earth's surface at different time and spatial scales, using examples to illustrate your points. You could also include information about the clues that geologists look for to determine the history of a rock formation, and how these clues can be used to understand the rock cycle. Finally, you could conclude by summarizing the key points and explaining the significance of the rock cycle in understanding the Earth's surface.
The United States uses only 10% of the world's energy.
True
False
Answers
Answer:
False
Explanation:
17%
the answer to your question is false
During a car accident on the NJ Turnpike, the airbags deploy. A 79 kg passenger traveling at 32 m/s makes impact with the airbag over a time of 0.25 seconds. What
was the impact force experienced by the passenger?
Answers
Hi there!
Recall that:
Impulse = Δ in momentum = mΔv
Impulse = Force · time
Begin by calculating the change in momentum, or impulse.
I = mΔv = m(vf - vi)
I = (79)(0 - 32) = -2528 Ns
Now, we can use the equation relating force and time to impulse.
I = Ft
Rearrange for time:
I/F = t
-2528/0.25 = -10112 N
**OR, if magnitude ⇒ |-10112| = 10112 N
What’s the meaning of physics??
Answers
Answer:the branch of science concerned with the nature and properties of matter and energy. The subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms.
Explanation:
A ball is projected at an angle of 53°. If the initial velocity is 48 meters/second, what is the vertical component of the velocity with which it was launched?
A.) 31 meters/second
B.) 38 meters/second
C.) 44 meters/second
D.) 55 meters/second
Answers
Answer:
The vertical component of the velocity can be found using the formula:
V₀y = V₀ * sin(θ)
where V₀ is the initial velocity, θ is the angle of projection, and V₀y is the vertical component of the velocity.
Substituting the given values, we have:
V₀y = 48 * sin(53°)
Using a calculator, we can evaluate sin(53°) to be approximately 0.799:
V₀y = 48 * 0.799
V₀y ≈ 38.352
Therefore, the vertical component of the velocity with which the ball was launched is approximately 38 meters/second, which corresponds to option B.
Answer:
B.) 38 meters/second
Explanation:
1. Calculate the electric field due to a single +1nC point charge at a distance of lm, 2m, and 3m
Answers
Answer:
Approximately \(9.0\; \rm N \cdot C^{-1}\) at \(1\; \rm m\) from this charge, pointing away from the point charge.Approximately \(2.2\; \rm N \cdot C^{-1}\) at \(\rm 2\; \rm m\) from this charge, pointing away from the point charge.Approximately \(1.0\; \rm N \cdot C^{-1}\) at \(3\; \rm m\) from this charge, pointing away from the point charge.Assumption: there is no object between this point charge and the observer.
Explanation:
The electric field of a point charge is inversely proportional to the square of the distance from that point charge.
Let \(k\) denote Coulomb's constant (\(k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-1}\).) Let the magnitude of that point charge be \(q\). At a distance of \(r\) from this charge, the electric field due to this charge would be:
\(\displaystyle E = \frac{k \cdot q}{r^{2}}\).
Convert the magnitude of the point charge in this question to standard units:
\(q = 1\; \rm nC = 10^{-9}\; \rm C\).
Apply that equation to find the magnitude of the electric field due to this point charge:
\(r = 1\; \rm m\):
\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(1\; \rm m)^{2}} \\ &\approx 9.0\; \rm N \cdot C^{-1}\end{aligned}\).
\(r = 2\; \rm m\):
\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(2\; \rm m)^{2}} \\ &\approx 2.2\; \rm N \cdot C^{-1}\end{aligned}\).
\(r = 3\; \rm m\):
\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(3\; \rm m)^{2}} \\ &\approx 1.0\; \rm N \cdot C^{-1}\end{aligned}\).
The direction of the electric field at a point is the same as the direction of a force from this field onto a positive point charge at this point.
Because the \((+1\; \rm nC)\) point charge here is positive, the electric field of this charge would repel other positive point charges. Hence, the electric field around this \((+1\; \rm nC)\!\) point charge at any point in the field would point away from this charge.
If Bill threw a ball straight up on the Moon (g=1.6 m/s2) with a starting velocity of 22m/s from a cliff and it fell past him and reached a velocity of 56m/s as it impacted the Moon’s surface below the cliff, how long was the flight time?
Answers
Answer:
Write two important of physical state
Which graph shows the change in velocity of an object in free fall?
Answers
Answer:
The graph of the velocity of an object in free fall would look like a straight line sloping downward. As the object falls, its velocity increases at a constant rate, so the graph of its velocity versus time will be a straight line with a negative slope. This is because acceleration due to gravity is a constant -9.8 meters per second squared, so the velocity of a free-falling object will increase by 9.8 meters per second every second.
Therefore, the graph that shows the change in velocity of an object in free fall is a straight line with a negative slope. Here is an example of such a graph:
Free Fall Velocity Graph
NEED HELP ON QUESTION ASAP! !
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Please can you explain what this paragraph is trying to say. Also what does it mean in the sentence 'the difference in charge across the battery provides push for current' and what is the difference in charge.
Here's paragraph I need to have a simple definition of:
A high waterfall is also like a large voltage. It will transfer a lot of energy to the water (charge), making the river flow very fast (a large current) the difference in height makes the river flow. In a circuit , the difference in charge across battery provides push for the current.
Answers
It is saying that a high waterfall has a lot of potential energy that can transfer to the water and cause it to flow very fast, which is similar to how a large voltage in an electric circuit can create a large current. The difference in height of the waterfall causes the water to flow, just as the difference in charge across a battery provides the push or force for the current to flow in a circuit.
In an electric circuit, voltage is the measure of the electric potential difference between two points, and it is measured in volts. Current is the flow of electric charge through the circuit, and it is measured in amperes or amps. The difference in charge across the battery refers to the potential difference or voltage between the positive and negative terminals of the battery, which creates an electric field and provides the force or push for the current to flow through the circuit.
a ball is thrown straight up into the air with an initial speed of 3.1 m/s. a. After 0.24 seconds what is the ball's velocity and b. what is it's acceleration?
Answers
a. The ball's velocity after 0.24 s is 0.75 m/s
b. The acceleration of the ball is given by the acceleration due to gravity
a. The ball's velocity can be calculated with the following equation:
\( v_{f} = v_{0} - gt \)
Where:
\( v_{f} \): is the final speed =?
\(v_{0}\): is the initial speed = 3.1 m/s
g: is the acceleration due to gravity = 9.81 m/s²
t: is the time = 0.24 s
The minus sign is because the acceleration is in the opposite direction (downward) of the motion of the ball (upward).
The final speed is:
\( v_{f} = v_{0} - gt = 3.1 m/s - 9.81 m/s^{2}*0.24 s = 0.75 m/s \)
Hence, the ball's velocity after 0.24 s is 0.75 m/s.
b. The acceleration of the ball is given by the acceleration due to gravity because the ball is thrown straight up (the motion of the ball is in the y-direction). The velocity of the ball in the x-direction is zero so the acceleration in the same direction is also zero.
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Write the differences between rest and motion
Answers
Answer:
Depends on position
Explanation:
Rest:
A body is said to be at rest, if it does not change its position with respect to its surroundings.
Motion:
A body is said to be in motion, if it changes its position with respect to its surroundings.
The state of rest and motion is relative
A robotic vehicle, or rover, is exploring the surface of Mars. The stationary Mars lander is the origin of coordinates, and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-coordinates that vary with time: = 2.0 − (0.25 / 2 ) 2 = (1.0/) + (0.025/ 3 ) 3 () ℎ ′ and distance from the lander at t=2.0s. (B) Find the rover’s displacement and average velocity vectors for the interval t=0.0 s to t=2.0s. (c) Find a general expression for the rover’s instantaneous velocity vector ⃗ . Express ⃗ at in component form and in terms of magnitude and direction
Answers
A) The rover’s coordinates and its distance from the lander at t = 2.0 s are; (1, 4) and 4.1 m
B) The rover’s displacement and average velocity vector during the interval are; s = (-1, 4) and v = (-0.5, 2) m/s
C) The magnitude and direction of the instantaneous velocity are; 2.24 m/s and 117°
What is the displacement and Velocity?
The rover's x and y coordinates are given as;
x = 2.0m − (0.25 m/s²)t²
y = (1.0m/s)t + (0.25 m/s³)t³
A) At t = 2 s, the rovers coordinates are;
x = 2.0m − (0.25 m/s²)2²
x = 1 m
y = (1.0m/s)2 + (0.25 m/s³)2³
y = 4 m
Distance from the lander is;
s = √[(1 - 2)² + 4²]
s = 4.1 m
B) Let us first find the distance coordinates for the interval t = 0.0 s to t = 2.0s. Thus;
s = r - r₀
s = (1 - 2), (4 - 0)
s = (-1, 4)
Thus, average velocity vector is;
v = ¹/₂s
v = ¹/₂(-1, 4)
v = (-0.5, 2) m/s
C) A general expression for the instantaneous velocity components is;
v_x = -0.5t
v_y = 1 - 0.75t²
Thus, v(2) is;
v_x = -0.5(2) = -1
v_y = 1 - 0.75(2)²
v_y = -2
Instantaneous velocity vector is; v = (-1, -2)
Magnitude of instantaneous Velocity = √(-1² + -2²) = 2.24 m/s
Direction = 180° - tan⁻¹(-2/-1) ≈ 117°
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If objects are traveling in opposite directions, what do you know about the signs of their momenta?
Answers
Explanation:
the impulse and momentum change on each object are equal in magnitude and opposite in directions. Thus the total momentum is reserved
They have no overall momentum at all. They are travelling in opposing directions yet having the same mass and velocity. Their momentum vectors add up to exactly zero when added together.
What If objects are travelling in opposite directions?When two objects collide, opposite-direction forces of equal magnitude are exerted to each item. When there are such pressures, it usually happens that one item speeds up and gains momentum, while the other object slows down (lose momentum).
Think about a situation where two similar objects are going in opposite directions at the same speed. It's noteworthy to note that despite both items moving, the momentum of the system as a whole is zero because the oppositely oriented vectors cancel out.
Therefore, Every object experiences an equal but opposite change in impulse and momentum. Thus, the entire momentum is held back.
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A team comprised of three people must run single file and leap onto an initially stationary, perfectly symmetrical, box-like, 4.0m long, 240kg sled (with uniform density) on frictionless ice (they begin to run before they encounter the ice). Their masses are 78 kg, 84 kg and 55 kg; their velocities are 19 m/s, 18 m/s and 16 m/s respectively; when the lead person is 3.0 m from the rear sss of the sled, the spacing between each person is 2.0 m.
(a) Determine the centre of mass (measured along the horizontal only) of the team/sled system when the lead person is 3.0 m from the rear of the sled.
(b) Determine the centre of mass of the team/sled system once all three team members have leapt onto the sled. (You may assume, once on board, the team members are equally spaced between the front and rear of the sled with one meter gaps between them.)
(c) Once the team is on board, the sled encounters a 30◦ incline above the horizontal for which the coefficient of kinetic friction is 0.313; what maximum displacement does the sled achieve measured along the surface of the incline?
Answers
how does the formation of bubbles while heating a liquid affect the coefficient of cubical expansion γ of a liquid?
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The formation of bubbles while heating a liquid can affect the coefficient of cubical expansion γ of the liquid. The coefficient of cubical expansion is a measure of how much a liquid expands when its temperature increases.
When bubbles form in a liquid, they create voids that decrease the effective volume of the liquid. As a result, the apparent expansion of the liquid is reduced, which decreases the value of γ.
The effect of bubble formation on the coefficient of cubical expansion depends on the size, number, and location of the bubbles. For example, if the bubbles are very small and evenly distributed throughout the liquid, the effect on γ may be negligible. On the other hand, if the bubbles are large and located near the surface of the liquid, they may significantly reduce the effective volume of the liquid and decrease the value of γ.
In addition, the formation of bubbles can also lead to other effects that affect the expansion of the liquid. For example, the vaporization of the liquid can cause a decrease in pressure, which can also affect the value of γ. Therefore, the formation of bubbles while heating a liquid can have complex effects on the coefficient of cubical expansion of the liquid, and the exact effect depends on various factors.
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Explain why the subatomic particle that is not a part of the nucleus stays bound to the atom. For full credit, you must include the charges of all subatomic particles in your explanation.
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The subatomic particle that is not a part of the nucleus which stays bound to the atom is the electron and is to maintain relative stability.
What is Electron?This is defined as a sub atomic particle which is negatively charged and revolve around the nucleus. In an atom the number of electron is usually equal to the number of the proton which is positively charged and present in the nucleus.
The electron not being in the nucleus is a a result of it wanting to maintain stability because its presence will produce a wavelength which will break the nucleus apart.
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03: Hook's law suggests that F is directly proportional to -x, how much true you have found this statement in your experiment? Explain any differences.
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Hooke's Law can be given as follows sometimes:
The restoring force of a spring is equal to the spring constant multiplied by the displacement from its normal position:
F = -kx
Where, F = Restoring force of a spring (Newtons, N)
k = Spring constant (N/m)
x = Displacement of the spring (m)
The negative sign relates to the direction of the applied force and by convention, the minus or negative sign is present in F = -kx. The restoring force F is directly proportional to the displacement (x), according to Hooke's law. When the spring is compressed, the displacement (x) is negative. It is zero when the spring is at its original length and positive when the spring is extended.
Practically, Hooke's Law is applicable only within a limited frame of reference, and through experimenting, this statement proves to be true. Because materials cannot be compressed beyond a certain size or expanded beyond a certain size without some permanent deformation or change of their original state.
The law only applies under some conditions such as a limited amount of force or deformation. Factually, many materials will noticeably deviate from Hooke's law even before those elastic limits are reached.
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What are the other important factors contributing to your physical fitness besides
just physical activity?
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Answer:
nutrition and rest
Explanation: Rest makes sure your not tired so that contributes and nutrition makes sure your healthy.
What doesn't H2O stand for
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Answer:
Water
Explanation:
A flywheel having constant angular acceleration requires 4.70 s to rotate through 164 rad . Its angular velocity at the end of this time is 101 rad/s . Find the angular velocity at the beginning of the 4.70 s interval. Find the angular acceleration of the flywheel.
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Answer:
A) -31.2 rad/s
B) 28.1 rad/s^2
Explanation:
Please help me in questions 4_5_6
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Yes, the force of the engine does work on the car as it accelerates the car forward. Work is defined as the transfer of energy that occurs when a force is applied over a distance, and in this case, the force of the engine is causing the car to move, so it is doing work.
How to explain the forceAs the car gains speed, its kinetic energy increases, which means it has more energy of motion. Kinetic energy is defined as one-half of the mass of an object times its velocity squared, so as the car's speed increases, so does its kinetic energy.
The gravitational potential energy of the car will remain constant as long as it stays on a horizontal road, assuming there is no change in elevation. Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, and since the car's position relative to the ground is not changing, its gravitational potential energy will remain constant.
The tractive force of the engine may change depending on the speed of the car and the resistance to motion that the car is encountering. As the car speeds up, the air resistance acting on the car will increase, which will require more force from the engine to maintain the same acceleration. Additionally, if the road surface is rough or there are inclines, the tractive force required to maintain the same speed or acceleration will also increase.
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Dillon wonders if he can block the radio waves released by the remote control used to operate his radio-controlled car. He covers the remote control in different materials and tries to use it to operate his car. He records his observations in the table below. Why does the remote control not work when covered with aluminum foil? A) because the aluminum foil blocks the batteries B) because the aluminum foil blocks the microwaves C) because the aluminum foil blocks the radio waves D) because the aluminum foil blocks the sound waves
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The remote control operated on Dillon's radio controlled car does not work when covered in aluminum foil because the aluminum foil blocks the radio waves. (option C)
Radio-controlled cars operate using radio waves, which are a type of electromagnetic radiation. When the remote control is covered with aluminum foil, the foil acts as a barrier that blocks the radio waves from reaching the car. This is why the remote control does not work when covered with aluminum foil.
It is important to note that radio waves are not the same as microwaves or sound waves. Microwaves are a type of electromagnetic radiation that is used for cooking food in microwave ovens, while sound waves are mechanical waves that travel through the air and are used for hearing. Therefore, options B and D are incorrect. Additionally, the aluminum foil does not block the batteries in the remote control, so option A is also incorrect.
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6 waves pass by a point in 2 seconds. What is the frequency?
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Answer:
A frequency by 2 seonds is 0.5 Hertz.
Explanation:
why do the scientific method is said to critical thinking
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If the size of an object is 5.0 cm, and the size of the image formed by a lens is 15.0 cm, what is the magnification of the system?
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The magnification of the system is 3.
To calculate the magnification, use the following system: magnification = the height of the photograph ÷ via the peak of the object. Plug your records into the formula and remedy. if your solution is more than 1, that means the photo is magnified. in case our solution is between zero and 1, the image is smaller than the object.
magnification = height of imagect/height of objest
= 15 cm/5 cm
= 3.
A magnification of one (plus or minus) approach that the image is the equal size as the object. If m has a value extra than 1 the photograph is greater than the item, and if an m with significantly much less than 1 way the photo is smaller than the item.
A negative magnification shows that the photograph is inverted. If the object is positioned in the direction of a converging lens than the focal period, the rays on the far side of the lens diverge. by means of extrapolating those traces returned (the red traces) the location of the digital image can be observed.
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