Answers
Givens.
• The mass of the ballon is 2.30 kg.
,• The height is 35.65 m.
First, find the final velocity when the balloon is at the bottom. Use a formula that relates height, initial speed, final speed, and gravity.
\(v^2_f=v^2_0+2gh\)Where
• v_0 = 0 because the balloon starts from rest.
,• Gravity is g = 9.8 m/s^2.
,• h = 35.65 m.
Use all these magnitudes to find the final velocity v_f
\(\begin{gathered} v^2_f=0^2+2(9.8\cdot\frac{m}{s^2})(35.65m) \\ v^2_f=698.74m^2 \\ v_f=\sqrt[]{698.74m^2} \\ v_f\approx26.4(\frac{m}{s}) \end{gathered}\)Once we have the velocity at the bottom, find the kinetic energy using its formula.
\(K=\frac{1}{2}mv^2\)Where m = 2.30 kg, and v = 26.4m.s.
\(\begin{gathered} K=\frac{1}{2}\cdot(2.30\operatorname{kg})\cdot(26.4(\frac{m}{s}))^2 \\ K=1.15\cdot696.96J \\ K=801.5J \end{gathered}\)Therefore, the kinetic energy at the bottom is 801.5 J.
Related Questions
You see a boat sitting at the end of a dock. Ten minutes later you see the same boat in a cove to the right of the dock. You did not see the boat move.
However, you know that the boat moved because its
relative to the dock changed.
Answers
The boat must have moved, despite not being seen to move, because its relative position to the dock has changed. This phenomenon is known as relative motion .
Everything is always in motion, but the way we perceive it depends on our frame of reference.
In this scenario, the dock was the frame of reference for the initial position of the boat. When the boat moved to the cove, its position relative to the dock changed, and the dock was no longer an appropriate frame of reference. The boat's motion is now relative to the cove instead.
It is important to note that relative motion depends on the chosen frame of reference. If we were to choose the boat as the frame of reference, then it would be the dock that appears to move, not the boat. This is because motion is always relative to a chosen frame of reference.
In conclusion, the boat must have moved because its position relative to the dock changed. The concept of relative motion reminds us that motion is always relative to a chosen frame of reference, and that the way we perceive motion depends on our chosen frame of reference.
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What will happen to the ice cubes in the two spoons? How will they be different?
Answers
Answer: the ice cube to melt more quickly on the metal block than on the wooden block.
Explanation:
Which undergoes the larger change in gravitational potential energy when dropped: a blue ball that is 20 feet above the ground, or a red one with the same mass that is 10 feet above the bottom of a nearby 50-foot-deep well?
Answers
A red one with the same mass that is 10 feet above the bottom of a nearby 50-foot-deep well will undergo a larger change in gravitational potential energy when dropped.
What is Gravitational potential energy?This is referred to as the energy an object which is possessed because of its position in a gravitational field.
Potential energy depends on an arbitrary zero point, not an absolute, like sea level which therefore explains why the red ball with the 10 foot drop into a well has more potential energy than the blue ball with only a 20 foot drop to Earth's surface.
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580 nm light shines on a double slit with d = 0.000125 m. What is the angle of the first bright interference maximum (m=1)?
PLS ANSWER
Answers
The angle of the first bright interference maximum (m=1) can be calculated using the equation: θ = sin-1(mλ/d).
What is equation ?An equation is a mathematical expression that uses symbols to describe a relationship between two or more variables. Equations are typically used to express relationships between physical quantities, such as forces and masses, or distances and times. They are also used to describe chemical reactions and other phenomena. Equations can be used to solve for unknown values, or to describe patterns or trends in data. Equations are typically written using algebraic notation, which includes variables, constants, and operators.
Substituting the given values, we get,θ = sin-1(1(580 × 10-9m)/(0.000125m)),θ = 0.00463 radians or 0.2637° .
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impulse is the product of a force and the blank during which that force acts
Answers
Calculate the young modulus in a cantilever depression method. The length is 1m which is suspended by a load of 150g. The depression is found to be 4cm. The thickness of the beam is 5mm and the breath is 3cm.(g=9.8)
Answers
The young's modulus in a cantilever will be 3.92 x 10¹⁰ N/m².
What is young's modulus?Young's modulus (E) is a material property that indicates how easily it can stretch and deform and is defined as the ratio of tensile stress () to tensile strain (). Where stress denotes the amount of force applied per unit area ( = F/A) and strain denotes the extension per unit length
Given that the length is 1m which is suspended by a load of 150g. The depression is found to be 4cm. The thickness of the beam is 5mm and the breath is 3cm.
The young's modulus will be calculated by the formula below,
Y = (4gl³) / (bd³) x ( M / y )
Y = ( 4 x 9.81 x 1³ x 0.150 ) / ( 0.03 x 0.005³ x 0.04 )
Y = 3.92 x 10¹⁰ N/m².
Therefore, young's modulus will be 3.92 x 10¹⁰ N/m².
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Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface
Answers
Answer:
(a) 4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface, (b) The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.
Explanation:
The complete statement is: "Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface, (b) How far has the rocket traveled when half the total work has occurred?"
(a) The work required to propel the rocket is given by the change in gravitational potential energy, whose expression derives is described below:
\(U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right]\)
Where:
\(U_{g,o}\), \(U_{g,f}\) - Initial and final gravitational potential energies, measured in joules.
\(m\), \(M\) - Masses of the rocket and planet Earth, measured in kilograms.
\(G\) - Universal gravitation constant, measured in newton-square meters per square kilogram.
\(r_{o}\), \(r_{f}\) - Initial and final distances of the rocket with respect to the center of the Earth, measured in meters.
The initial distance and rocket mass are converted to meters and kilograms, respectively:
\(r_{o} = (4000\,mi)\cdot \left(1609.34\,\frac{m}{mi} \right)\)
\(r_{o} = 6,437,360\,m\)
\(m = (7\,ton)\cdot \left(1000\,\frac{kg}{ton} \right)\)
\(m = 7000\,kg\)
Given that \(m = 7000\,kg\), \(M = 5.972\times 10^{24}\,kg\), \(G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}\), \(r_{o} = 6,437,360\,m\) and \(r_{f} \rightarrow +\infty\), the work equation is reduced to this form:
\(U_{g,f} - U_{g,o} = \frac{G\cdot m \cdot M}{r_{o}}\)
\(U_{g,f} - U_{g,o} = \frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)}{6,437,360\,m}\)
\(U_{g,f} - U_{g,o} = 4.334\times 10^{11}\,J\)
4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface.
(b) The needed change in gravitational potential energy is:
\(U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J\)
The expression for the change in gravitational potential energy is now modified by clearing the final distance with respect to the center of Earth:
\(U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right]\)
\(\frac{U_{g,o}-U_{g,f}}{G\cdot M \cdot m} = \frac{1}{r_{f}} - \frac{1}{r_{o}}\)
\(\frac{1}{r_{f}} = \frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m}\)
\(r_{f} = \left(\frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m} \right)^{-1}\)
If \(m = 7000\,kg\), \(M = 5.972\times 10^{24}\,kg\), \(G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}\), \(r_{o} = 6,437,360\,m\) and \(U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J\), then:
\(r_{f} = \left[\frac{1}{6,437,360\,m}-\frac{2.167\times 10^{11}\,J}{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)} \right]^{-1}\)
\(r_{f} \approx 12,874,502.49\,m\)
The final distance with respect to the center of the Earth in miles is:
\(r_{f} = (12,874,502.49\,m)\cdot \left(\frac{1}{1609.34}\,\frac{mi}{m} \right)\)
\(r_{f} = 7999.865\,mi\)
The distance travelled by the rocket is: (\(r_{f} = 7999.865\,mi\), \(r_{o} = 4000\,mi\))
\(\Delta r = r_{f}-r_{o}\)
\(\Delta r = 7999.865\,mi - 4000\,mi\)
\(\Delta r = 3999.865\,mi\)
The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.
A slide with an image 4cm×2cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen 1.1 m from the slide. the size of the image on the screen is
Answers
Answer:
44cm x 22cm
Explanation:
u= 10 cm
v= 1.1 cm
m=v/u= 1.1/10
m=11
hence the size of the image.
I need the answer to this question
Answers
Answer:
I don't see anything so you should repost
Some cats can be trained to jump from one location to another and perform other tricks. Kit the cat is going to jump through a hoop. He begins on a wicker cabinet at a height of 1.765 m above the floor and jumps through the center of a vertical hoop, reaching a peak height 3.130 m above the floor. (Assume the center of the hoop is at the peak height of the jump. Assume that +x axis is in the direction of the hoop from the cabinet and +y axis is up. Assume g = 9.81 m/s2.)
(a) With what initial velocity did Kit leave the cabinet if the hoop is at a horizontal distance of 1.560 m from the cabinet?
v_0 = m/s
(b) If Kit lands on a bed at a horizontal distance of 3.582 m from the cabinet, how high above the ground is the bed?
m
Answers
Answer:
a. the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal
b. 0.847 m
Explanation:
a. Using v² = u² + 2as, we find the initial vertical velocity of the cat. Now at the peak height, v = final velocity = 0, u = initial velocity and a = -g = 9.8 m/s², s vertical distance travelled by the cat from its position on the cabinet = Δy = 3.130 m - 1.765 m = 1.365 m.
Substituting these variables into the equation, we have
0² = u² + 2(-9.8m/s²) × 1.365 m
-u² = -26.754 m²/s²
u = √26.754 m²/s²
u = 5.17 m/s
To find its initial horizontal velocity, u₁ we first find the time t it takes to reach the peak height from
v = u + at. where the variables mean the same as above.
substituting the values, we have
0 = 5.17 m/s +(-9.8m/s²)t
-5.17 m/s = -9.8m/s²t
t = -5.17 m/s ÷ (-9.8m/s²)
= 0.53 s
Now, the horizontal distance d = u₁t = 1.560 m
u₁ = d/t = 1.560 m/0.53 s = 2.96 m/s
So, the initial velocity of the cat is V = √(u² + u₁²)
= √((5.17 m/s)² + (2.96 m/s)²)
= √(26.729(m/s)² + 8.762(m/s)²)
= √(35.491 (m/s)²)
= 5.95 m/s
its direction θ = tan⁻¹(5.17 m/s ÷ 2.96 m/s) = 60.2°
So, the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal
(b)
First, we find the time t' it takes the cat to land on the bed from d' = u₁t'
where d' = horizontal distance of cabinet from bed = 3.582 m
u₁ = horizontal velocity = 2.96 m/s
t' = d'/u₁
= 3.582 m/2.96 m/s
= 1.21 s
The vertical between the bed and cabinet which is the vertical distance moved by the cat is gotten from Δy = ut' +1/2at'²
substituting u = initial vertical velocity = 5.17 m/s, t' = 1.21 s and a = -g = -9.8 m/s² into Δy, we have
Δy = ut' +1/2at'² = 5.17 m/s × 1.21 s +1/2(- 9.8 m/s²) × (1.21 s)² = 6.256 - 7.174 = -0.918 m
Δy = y₂ - y₁
Since our initial position is the position of the cabinet above the ground = y₁ = 1.765 m
y₂ = position of bed above ground.
Δy = y₂ - y₁ = -0.918 m
y₂ - 1.765 m = -0.918 m
y₂ = 1.765 m - 0.918 m
= 0.847 m
Convert 15000kg/m cube
into gm/cm cube
please write the process also
Answers
Answer:
15000 Kilograms/Cubic Meters (kg/m3) = 15 Grams/Cubic Centimeters (g/cm3)
Explanation:
1 g/cm3 is equal to 1000 kilogram/cubic meter. To convert 100 gram into kg then divide it by 1000 i.e. 100/1000 = 0.1 kg. To convert any value of gm/cm3 into kg/m3 then multiply it by 1000.
15000 kg / m^3 =
15000 × 10^3 g / m^3 =
15000 × 10^3 × 10^3 mg / m^3 =
15 × 10^9 mg / m^3 =
15 × 10^9 × 10^(-3) mg / dm^3 =
15 × 10^9 × 10^(-3) × 10^(-3) mg / cm^3 =
15 × 10^9 × 10^(-6) mg / cm^3 =
15 × 10^( 9 - 6 ) mg / cm^3 =
15 × 10^3 mg / cm^3 =
15000 mg / cm^3 =
Look : We found the exact thing we had ...
WoW ...
We got a point ;
Remember from now on :
kg / m^3 = mg / cm^3
ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements. If a vector that is 3 cm long represents 30 km/h, what velocity does a 5 cm long vector which is drawn using the same scale represent?
50kmh
60kmh
100kmh
Answers
Can someone please help me with this worksheet. It’s due today and I have no idea what to do.
Answers
why do we preserve food?
Answers
Answer:
The primary objective of food preservation is to prevent food spoilage until it can be consumed. Gardens often produce too much food at one time—more than can be eaten before spoilage sets in. Preserving food also offers the opportunity to have a wide variety of foods year-round. It's economic.
Explanation:
A 30.0 kg box hangs from a rope. What is the tension (in N) in the rope if the box has an initial velocity of 2.0 m/s and is slowing down at 2.0 m/s2?
Answers
Answer:
360 N
Explanation:
m = 30kg u = 2 m/s a = -2m/s/s
Since the object has an initial velocity of 2 m/s and acceleration of -2 m/s/s
the object will come to rest in 1 second but the force applied in that one second can be calculated by:
F = ma
F = 30 * -2
F = -60 N (the negative sign tells us that the force is acting downwards)
Now, calculating the force applied on the box due to gravity
letting g = -10m/s/s
F = ma
F = 30 * -10
F = -300 N (the negative sign tells us that the force is acting downwards)
Now, calculating the total downward force:
-300 + (-60) = -360 N
Hence, a downward force of 360 N is being applied on the box and since the box did not disconnect from the rope, the rope applied the same amount of force in the opposite direction
Therefore tension on the force = 360 N
An elevator is moving upward with a speed of 9.2 m/s. 3.8 seconds later, the elevator is still moving upward, but its speed has increased to 15.8 m/s. What is the average acceleration, in m/s2, of the elevator during the 3.8-s interval? Please round your answer to one decimal place.
Answers
From the calculations, the acceleration of the elevator is obtained as 1.5m/s^2
What is acceleration?The term acceleration has to do with the rate of change of velocity with time.
Given that;
u = 9.2 m/s
v = 15.8 m/s
t = 3.8
a = ?
The v = u + at
a = v - u/t
a = 15 - 9.2/3.8
a = 1.5m/s^2
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A tow truck is pulling a car out of a ditch by means of a steel cable (Y-2.0 x 10¹1 N/m²) that is 9.86 m long and has a radius of 0.413 cm.
When the car just begins to move, the tension in the cable is 922 N. How much has the cable stretched?
Answers
The extension of the cable is 8.48 x 10⁻⁴ N/m².
What is the extension of the cable?
The extension of the cable is calculated by applying the formula for Young's modulus as shown below.
ΔL = ( FL₀ ) / ( AY )
where;
Y is the Young's modulusF is the tension in the cableL₀ is the original length of the cableA is the area of the cableThe extension of the cable is calculated as follows;
ΔL = ( 922 x 9.86 ) / ( π x 0.00413² x 2 x 10¹¹ )
ΔL = 8.48 x 10⁻⁴ N/m²
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Are Quantum Physics, Quantum mechanics,Quantum Engagement same?
or, Do they branch of each others
Answers
Answer:
The topic of quantum entanglement is at the heart of the disparity between classical and quantum physics: entanglement is a primary feature of quantum mechanics lacking in classical mechanics. ... In the case of entangled particles, such a measurement will affect the entangled system as a wholeExplanation:
Answer:
quantum entanglement is thought to be one of the trickiest concepts in science, but the core issues are simple. And once understood, entanglement opens up a richer understanding of concepts such as the “many worlds” of quantum theory.
Explanation:
. Which of the statement below about the magnetic
properties of iron and steel is correct?
A. Both iron and steel are equally magnetized and
demagnetized
B. Iron is more easilymagnetized and demagnetized
than steel.
C. Steel is more easily magnetized and demagnetized
than iron
D. Iron is more easily magnetized than steel but steel is
more easily demagnetized than iron.
Answers
Answer:
D. Iron is more easily magnetized than steel but steel is more easily demagnetized than iron.
7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.
Answers
The tension in the strings are 31.47 and 19.25 N respectively.
Mass of the block, m = 3 kg
From the figure, consider the vertical components,
T₁ sin45° + T₂ sin30° = mg
(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4
Also, consider the horizontal components,
T₁ cos45° = T₂ cos30°
T₁/√2 = T₂ x√3/2
T₁ = T₂ x √3/2 x √2
So,
T₁ = 0.612T₂
Applying in the first equation,
(T₁/√2) + (T₂/2) = 29.4
(0.612T₂/1.414) + 0.5T₂ = 29.4
0.434 T₂ + 0.5 T₂ = 29.4
0.934 T₂ = 29.4
Therefore, the tension,
T₂ = 29.4/0.934
T₂ = 31.47 N
So, the tension,
T₁ = 0.612 T₂
T₁ = 0.612 x 31.47
T₁ = 19.25 N
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Urgent!!! big brain needed!
A 66.4 kg person sits on the right end of a seesaw, 2.35 m from the fulcrum. On the other end, how far from the fulcrum should a 84.2 kg person sit to balance the seesaw?
1.92 m
1.85 m
2.08 m
1.61 m
Answers
1.85 m if On the right side of a seesaw, 2.35 meters from the fulcrum, a 66.4 kg individual is seated. How far away from the fulcrum on the opposite end should an 84.2 kg.
The correct statement is B.
What is a fulcrum with an example?For those who are unsure of what a fulcrum is, The answer to the question a fulcrum is the fixed point of a lever. A seesaw, for instance, is a lever with a fulcrum with in middle. The fulcrum of a seesaw is the stationary central component.
What are a pivot and a fulcrum?• The lever's pivot or turning point, or fulcrum. • The load, or the items you are attempting to transport. • The force, or the energy expended in moving the weight. With the help of a lever, you can move a heavy object with little effort. A lever is easier to move the farther away the effort (or force) comes from the fulcrum.
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What are the action and linking verb in: are you a spoiled child?
Answers
Explain/Describe how atoms in domains determine whether a material is magnetic or not. (Please help this is due today)
Answers
Answer:
In a material, the magnetic behavior depends on the alignment of magnetic moments of the atoms. Magnetic moments are generated by the motion of the electrons in the atoms. When the magnetic moments of atoms in a material are aligned in a specific pattern, it creates a magnetic field which results in the material being magnetic.
In many materials, the magnetic behavior arises due to the alignment of magnetic domains, which are regions of atoms with magnetic moments aligned in the same direction. When many domains with aligned magnetic moments are present in a material, the material becomes magnetic.
The magnetic behavior of a material depends on the number of electrons and the arrangement of those electrons in the atoms. In particular, for an atom to have a magnetic moment, it must have unpaired electrons, meaning electrons that are not paired with another electron with the opposite spin. When these unpaired electrons in the atoms are aligned, they generate a magnetic moment. If all electrons are paired, there will not be a net magnetic moment, so the material will not be magnetic.
So, in summary, the magnetic behavior of a material is determined by the alignment of magnetic moments of atoms. When the magnetic moments of many atoms in a material align in the same direction, it creates a magnetic field, leading to a material being magnetic. This alignment is usually present in magnetic domains consisting of atoms with unpaired electrons.
Please Help!!
A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?
Answers
Answer:
h = 219.96 m
Explanation:
Speed of the stone with which it was thrown horizontally, v = 10 m/s
We need to find the height of the cliff if the stone’s trajectory time from the top of the hill to the bottom to be 6.7s.
It means we need to find the distance covered by the stone. As the horizontal speed of the stone is given , it means there is no vertical motion in the stone,u'=0
Using second equation of motion,
\(d=u't+\dfrac{1}{2}at^2\)
Put u'=0 and a=g
So,
\(d=\dfrac{1}{2}gt^2\\\\\text{Putting all the values to find d}\\\\d=\dfrac{1}{2}\times 9.8\times 6.7^2\\\\d=219.96\ m\)
So, the height of the cliff is 219.96 m.
What is the component of the trucks weight parallel to the hill
Answers
Answer:
Below
Explanation:
It will be m * g * sin (angle of hill) You will need to put in the numbers for 'm' and sin (angle) and g = 9.81 m/s^2
Car A is moving at a speed of 45km/h towards car B which is moving at a speed of 55km/h. if the two car were initially separated at a distance of 150km, determine how long it will take the two cars to meet?
Answers
The time taken by the cars to meet is 5.4 x 10³ s.
Speed of car A, v₁ = 45 km/h = 12.5 m/s
Speed of car B, v₂ = 55 km/h = 15.27 m/s
Distance between the cars, d = 150 km = 15 x 10⁴m
The expression for the time taken by the cars to meet can be given as,
Time = Distance/Average speed
t = d/(v₁ + v₂)
Applying the values of d, v₁ and v₂.
t = 15 x 10⁴/(12.5 + 15.27)
t = 15 x 10⁴/27.77
t = 5.4 x 10³ s
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a fluid that becomes more dense will sink or rise?
Answers
It depends on the liquid the fluid is being put in. If the liquid is more dense than the fluid then it will sink but if the fluid is denser than the liquid then it will float.
if the fluid becomes more dense as it is put in the liquid it can sink as long as it reaches a high spenough density to be denser than this liquid it is in.
A flat uniform circular disk (radius = 5.44 m, mass = 150 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A 47.0-kg person, standing 1.54 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.80 m/s relative to the ground. Find the resulting angular speed (in rad/s) of the disk.
Answers
The resulting angular speed of the flat, uniform circular disk is 0.237 rad/s.
What is angular momentum?Angular momentum is a fundamental concept in physics that describes the rotational motion of an object around an axis. It is defined as the product of the moment of inertia of an object and its angular velocity with respect to a chosen axis.
We can use conservation of angular momentum to solve this problem. The initial angular momentum of the disk is zero because it is stationary. The final angular momentum of the system (disk + person) is:
L = Iω
where I is the moment of inertia of the disk and person about the axis of rotation, and ω is the resulting angular speed of the disk.
The moment of inertia of the disk about its axis is:
I_disk = (1/2)mr²
where the disk's radius is r and its mass is m. Substituting the given values, we get:
I_disk = (1/2)(150 kg)(5.44 m)² = 2226.24 kg·m²
The moment of inertia of the person about the axis can be approximated as:
I_person = mr²
where r is the distance from the axis to the person. Substituting the given values, we get:
I_person = (47.0 kg)(1.54 m)² = 109.64 kg·m²
The total moment of inertia of the system is:
I = I_disk + I_person = 2226.24 kg·m² + 109.64 kg·m² = 2335.88 kg·m²
The final angular momentum of the system is:
L = Iω
where ω is the resulting angular speed of the disk. Substituting the given values, we get:
(2335.88 kg·m²)ω = (197.64 kg·m²/s)(2.80 m/s)
Solving for ω, we get:
ω = (197.64 kg·m²/s)(2.80 m/s) / (2335.88 kg·m²) = 0.237 rad/s.
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A marine weather station reports waves along the shore that are 2 meters high, 8 meters long, and reach the station 8 seconds apart. Determine the speed of these waves.
Answers
One full weather station reports wave passing a spot every second is equivalent to a frequency of 1 Hz. The frequency is 48=12=0.5 Hz if 4 waves pass a spot in 8 seconds.
What is the frequency of a 30-second ocean wave that hits the shore?The number of cycles that make up a time unit is the frequency. The frequency of a wave with a 30-second period is therefore 1 30 = 0.033 cycles per second, or 0.033 Hertz (Hz).
Which of the following statements most accurately sums up how a wave crosses a border and enters a new medium?A wave's speed and wavelength vary as it crosses a border and enters a new medium, but its frequency doesn't.
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The results for one patient show that the blood in the aorta begins at a speed of 0.10 m/s and undergoes constant acceleration for 38 ms, reaching a peak speed of 1.29 m/s. (a) What is the acceleration reflected in these data? (b) How far does the blood travel during this period?
Answers
Answer:
s ≈ 0.0603 metersExplanation:
(a) To find the acceleration, we can use the equation of motion:
v = u + at
Where:
v = final velocity = 1.29 m/s
u = initial velocity = 0.10 m/s
a = acceleration (unknown)
t = time = 38 ms = 0.038 s
Rearranging the equation, we have:
a = (v - u) / t
Plugging in the values, we get:
a = (1.29 - 0.10) / 0.038
a = 1.19 / 0.038
a ≈ 31.32 m/s²
Therefore, the acceleration reflected in the data is approximately 31.32 m/s².
(b) To find the distance traveled, we can use the equation of motion:
s = ut + (1/2)at²
Where:
s = distance traveled (unknown)
u = initial velocity = 0.10 m/s
t = time = 38 ms = 0.038 s
a = acceleration = 31.32 m/s²
Plugging in the values, we get:
s = (0.10 × 0.038) + (0.5 × 31.32 × 0.038²)
s ≈ 0.0038 + 0.0565
s ≈ 0.0603 meters
Therefore, the blood travels approximately 0.0603 meters during this period.
Answer:
0.002 m
Explanation:
We can use the equations of motion to solve this problem.
(a) The initial velocity of the blood is u = 0.10 m/s, the final velocity is v = 1.29 m/s, the time taken is t = 38 ms = 0.038 s, and the acceleration is a (which is what we want to find). The equation that relates these quantities is:
v = u + at
Rearranging this equation, we get:
a = (v - u) / t = (1.29 - 0.10) / 0.038 = 33.68 m/s^2
Therefore, the acceleration of the blood is 33.68 m/s^2.
(b) To find the distance traveled by the blood during this period, we can use another equation of motion:
s = ut + (1/2)at^2
where s is the distance traveled. Substituting the values we have:
s = (0.10)(0.038) + (1/2)(33.68)(0.038)^2 = 0.002 m
Therefore, the blood travels a distance of 0.002 m during this period.