The magnitude of the force on the 1.0 nC charge is 3.8 N. The direction of the force on the 1.0 nC charge is towards the 2.0 nC charge, at an angle of 60°.
What is magnitude ?Magnitude is a measure of the size or strength of something. It can refer to physical quantities such as size, temperature, mass, or energy, as well as abstract qualities like loudness, brightness, or intensity. Magnitude is usually expressed as a number on a scale or as a unit of measurement, such as degrees or decibels. Magnitude can also refer to the importance or significance of an event or situation. For example, a hurricane may be said to have a magnitude of 5, or a financial crisis may have a magnitude of 10. Magnitude is an important concept in many scientific fields, such as physics and astronomy, where it is used to measure the size or power of an object or phenomenon.
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PLEASE HELP!! :)
Which of the following options would increase the electric force the most?
a. tripling the charge on one particle
b. changing the sign of one of the particles.
c. doubling the charge on one particle
d. doubling the charge on both particles
Calculate the quantity of heat energy which must be transferred to 2.25 kg of brass to raise its temperature from 20°C to 240°C if the specific heat of brass is 394 J/kgK.
The quantity of heat energy that must be transferred to 2.25 kg of brass to raise its temperature from 20 °C to 240 °C is 195030 J
How do i determine the quantity of heat energy?First, we shall list out the given parameters from the question. This is shown below:
Mass of brass (M) = 2.25 Kg Initial temperature of brass (T₁) = 20 °CFinal temperature of brass (T₂) = 240 °CChange in temperature of brass (ΔT) = 240 - 20 = 220 °CSpecific heat capacity of brass (C) = 394 J/kgKQuantity of heat energy (Q) =?The quantity of heat energy that must be transferred can be obtained as follow:
Q = MCΔT
= 2.25 × 394 × 220
= 195030 J
Thus, we can conclude quantity of heat energy that must be transferred is 195030 J
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b. Density and relative density.
Explanation:
density, mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre.
The relative density of a substance is defined as the ratio of the density of that substance to the density of water at 4oC. It is also defined as the ratio of the mass of substance to the mass of equal volume of water at 4oC. i.e., R.D. = Mass of the substance / mass of an equal volume of water at 4oC
An object of height 5cm is placed 20cm in front of pin hole camera from a cubic box of side 6cm . Determine the height of the image formed
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?
Answer:
5N
Explanation:
We have a simple problem of momentum here.
ΔMomentum= mΔv= FΔt
Solve for F
mΔv/Δt=F
Plug in givens
1*(2-1.5)/0.1=F
F=5N
The amount of force that the wall imparts on the ball is 5.0N
According to Newton's second law, the formula for calculating the force applied is expressed as:
\(F=ma\)
m is the mass of the object
a is the acceleration of the object
Since acceleration is the change in velocity of an object, hence \(a=\frac{\triangle v}{t}\)
The applied force formula becomes \(F=\frac{m\triangle v}{t}\)
Given the following parameters
m = 1.0kg
\(\triangle v=2.0-1.5\\\triangle v=0.5m/s\)
t = 0.1sec
Substitute the given parameter into the formula
\(F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N\)
Hence the amount of force that the wall imparts on the ball is 5.0N
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A beaker is filled with water to the rim. Gently you place a toy duck in the beaker and some of the water spills out. The duck floats. The weight of the beaker with the duck floating in it is Group of answer choices less than its weight before adding the duck. the same as its weight before adding the duck. greater or less than its weight before adding the duck, depending on the weight of the duck. greater than its weight before adding the duck.
Answer:
Explanation:
The same as its weight before adding the duck
True or False: For a longitudinal wave, the wavelength is the distance between compressions.
Answer:
false....
Explanation:
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Two dwarf galaxies revolve in circles around a normal large galaxy. Dwarf Galaxy A is 70,000 light years away from the large galaxy, and Dwarf Galaxy B is 500,000 ly away from the large galaxy. Which of the two galaxies, A or B, takes the longest time to orbit the large galaxy and what law tell us so
Answer:
B
Explanation:
From Kepler's Third Law, T^2 is proportional to r^3.
Now r of B is larger than r of A,
so T of B is larger than T of A.
Then B takes the longest time.
Dwarf Galaxy B takes the longest time to orbit the large galaxy. This can be explained using Kepler's Third Law of Planetary Motion, also known as the Law of Periods.
Kepler's Third Law states that the square of the orbital period (T) of a celestial body is directly proportional to the cube of the semi-major axis (a) of its elliptical orbit. Mathematically, it can be expressed as:
\(\[ T^2 \propto a^3 \]\)
Where:
T is the orbital period (time taken for one complete orbit), and
a is the semi-major axis of the elliptical orbit.
In the case of dwarf galaxies A and B revolving around the large galaxy, we can use the distances from the large galaxy to each of the dwarf galaxies as the semi-major axes (a) of their orbits.
Let's compare the values of \(\( a^3 \)\) for the two dwarf galaxies:
For Dwarf Galaxy A:
\(\( a_{\text{A}} = 70,000 \)\) light years
For Dwarf Galaxy B:
\(\( a_{\text{B}} = 500,000 \)\) light years
Now, let's calculate \(\( a^3 \)\) for each galaxy:
For Dwarf Galaxy A:
\(\( a_{\text{A}}^3 = (70,000)^3 = 343,000,000,000 \)\) cubic light years
For Dwarf Galaxy B:
\(\( a_{\text{B}}^3 = (500,000)^3 = 125,000,000,000,000,000 \)\) cubic light years
Since \(\( a_{\text{B}}^3 \)\) is significantly larger than \(\( a_{\text{A}}^3 \)\), it means that Dwarf Galaxy B has a larger a value and therefore a longer orbital period according to Kepler's Third Law. In other words, Dwarf Galaxy B takes the longest time to orbit the large galaxy.
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A student measures that 81,500 J of thermal energy were added to 0.5 kg of water. If the specific heat of water is 4,184 J/kg 0C, what was its change of temperature?
Answer:
\(\Delta T=38.95^{\circ} C\)
Explanation:
Given that,
Heat measured, Q = 81500 J
Mass of water, m = 0.5 kg
The specific heat of water is 4,184 J/kg °C
We need to find the change in temperature. The heat measured is given by :
\(Q=mc\Delta T\)
Where
\(\Delta T\) is the change in temperature
\(\Delat T=\dfrac{Q}{mc}\\\\\Delat T=\dfrac{81500}{0.5\times 4184 }\\\\\Delta T=38.95^{\circ} C\)
So, the change in temperature is \(38.95^{\circ} C\).
A ray of light traveling in water hits a glass surface. The index of refraction of the water is 1.33, and that of the glass is 1.50. At what angle with the plane of the surface must the incident ray strike the glass in order that the polarization of the reflected ray is the greatest
Answer:
\(\mu=41.5\textdegree\)
Explanation:
From the question we are told that:
Water index of refraction \(i_w=1.33\)
Glass index of refraction \(i_g=1.50\)
Generally the equation for Brewster's law is mathematically given by
\(\theta=tan^{-1}(\frac{i_g}{i_w})\)
\(\theta=tan^{-1}(\frac{1.50}{1.33})\)
\(\theta=48.44 \textdegree\)
Therefore Angle of incident to plane \mu (normal at 90 degree to the surface)
\(\mu=90\textdegree-\theta\)
\(\mu=90\textdegree-48.44\textdegree\)
\(\mu=41.5\textdegree\)
A golf ball has a mass of 0.046 kg. The final velocity of the ball
after being struck by a golf club is 50 ms-1. The golf club is in
contact with the ball for a time of 1.3 ms. Calculate the average
force exerted by the golf club on the ball.
The average force exerted by the golf club on the golf ball is equal to 1769.2N.
What is force?Force acting on an object can be described as the influence that changes the state of the object of motion or rest. The S.I. unit of force is Newton (N) and force is a vector parameter.
The mathematical formula of the 2nd law of motion for force is written as follows:
F = ma
Given the mass of the ball, m= 0.046 Kg
The golf club was in contact with the ball for time, t = 1.3 ms = 1.3 ×10⁻³s
The speed of the golf ball, v = 50 m/s
We know that, equation of motion: \(v = u + at\)
50 = 0 + a (1.3 ×10⁻³)
a = 38.46 ×10³ m/s²
The average force on the golf ball will be:
F = ma = 0.046 ×38.46 ×10³
F = 1769.2 N
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Nowton's third law refers to 'action reaction forces*. These forces are
always:
equal in magnitude but opposite in direction
A 6.47 mm high firefly sits on the axis of, and 12.3 cm in front of, the thin lens A, whose focal length is 5.77 cm. Behind lens A there is another thin lens, lens B, with a focal length of 20.9 cm. The two lenses share a common axis and are 58.1 cm apart.What is the height of this image? Express the answer as a positive number.
A 6.47 mm high firefly sits on the axis of, and 12.3 cm in front of, the thin lens A, whose focal length is 5.77 cm, the height of the image is 2.61 mm.
We can use the thin lens equation to find the image distance and then use the magnification equation to find the height of the image.
Let's call the distance between the firefly and lens A "d1", the distance between the lenses "d2", the image distance from lens B "d3", and the height of the firefly "h1".
Using the thin lens equation for lens A:
1/fA = 1/d1 + 1/d3
Since the firefly is very small, we can assume that the rays of light from it are parallel to the axis of the lenses, so d1 = 12.3 cm.
Solving for d3, we get:
1/d3 = 1/fA - 1/d1
1/d3 = 1/5.77 cm - 1/12.3 cm
d3 = -23.46 cm
The negative value for d3 indicates that the image is formed on the same side of lens B as the firefly, which means it is a virtual image.
Now we can use the magnification equation:
m = -d3/d2
where m is the magnification of the image. The negative sign indicates that the image is inverted.
Using the distance between the lenses, d2 = 58.1 cm, we get:
m = -(-23.46 cm) / 58.1 cm
m = 0.403
This tells us that the image is smaller than the firefly, and its height is:
h2 = m * h1
h2 = 0.403 * 6.47 mm
h2 = 2.61 mm
Therefore, the height of the image is 2.61 mm.
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wave length of sodium light is 5893A express in nanometre
Answer:
589.3 nm
Explanation:
The wavelength of a sodium light is 5893 A.
We need to find the wavelength in nm.
\(5893\ A=5893\times 10^{-10}\ m\)
Also, \(1\ nm=10^{-9}\ m\)
\(5893\ A=5893\times 10^{-9}\times 10^{-1}\ m\\\\=5893\times 10^{-1}\ nm\\\\=589.3\ nm\)
So, the wavelength of sodium light is 589.3 nm.
A cell of inter resistance of 0.5 ohm is connected to coil of resistance 4 ohm and 8 ohm joined in parallel.If there is current of 2A in 8 ohm,find the emf of the cell.
A cell of inter resistance of 0.5 ohm is connected to coil of resistance 4 ohm and 8 ohm joined in parallel.If there is current of 2A in 8 ohm, the electromotive force (emf) of the cell is approximately 14.5 volts.
To find the emf of the cell, we can apply Ohm's Law and Kirchhoff's laws to analyze the circuit.
Given:
Resistance of the coil, R1 = 4 ohm
Resistance of the other resistor, R2 = 8 ohm
Current passing through the 8-ohm resistor, I = 2A
First, let's analyze the parallel combination of the 4-ohm and 8-ohm resistors.
The total resistance of two resistors in parallel can be calculated using the formula:
1/Rp = 1/R1 + 1/R2
Substituting the given values, we have:
1/Rp = 1/4 + 1/8
1/Rp = 2/8 + 1/8
1/Rp = 3/8
Rp = 8/3 ohm
Now, let's consider the total resistance in the circuit, which includes the internal resistance of the cell (0.5 ohm) and the parallel combination of the resistors (8/3 ohm).
R_total = R_internal + Rp
R_total = 0.5 + 8/3
R_total = 1.833 ohm
Now, we can find the emf of the cell using Ohm's Law:
emf = I * R_total
emf = 2 * 1.833
emf ≈ 3.667 volts
Therefore, the emf of the cell is approximately 3.667 volts.
However, it is worth noting that the given current of 2A passing through the 8-ohm resistor does not affect the emf calculation since the emf of the cell is independent of the current in the circuit.
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What does this circle graph tell you about water on Earth? (2 points)
a pie graph with a big blue section covering seventy one percent and small gray section covering twenty nine percent with a key indicating that blue is water and gray is land
Fresh water covers 71 percent of Earth's surface.
Oceans covers 71 percent of Earth's surface.
Salt water covers 71 percent of Earth's surface.
Water covers 71 percent of Earth's surface.
Answer:
ocean covers 71 percent of the earth
Answer:
the ocean covers 71 percent of Earth's surface.
Explanation:
Q:- EXPLAIN WHY THE PLANETS DO NOT TWINKLE.
planets shine more stedily because they are closer to earth and so appear as pinpoints but as tiny disks in our sky.
A student builds an electromagnet using a battery, an iron nail, and some insulated wire. The wire is wrapped around the nail 50 times, then connected to the
battery. What changes could the student make to increase the strength of the electromagnet?
1. wrap more wire around the nail
2. increase the voltage of the battery
3. use a switch to turn the electromagnetic on and off
4. increase the size of the nail
A. 1 and 2
B. 2 and 4
C. 1,2 and 4
D. 1,2,and 3
Answer:
a
Explanation:
its right
To increase the electromagnet's strength, the student must enlarge the nail, wrap more wire around the nail, and increase the voltage of the battery. Option c is the correct answer.
How does an electromagnet work?The mechanics of an electromagnet are not overly complicated. It is accomplished by wrapping a length of conductive wire, usually copper, around a metal object.Until electricity is introduced, this appears to be nothing more than a jumbled collection of parts, similar to Frankenstein's monster. An electromagnet, on the other hand, does not require a storm to be activated. A current is introduced and flows through the wire, which can come from a battery or another source of power.As a result of the magnetic field created around the coiled wire, the metal is magnetized as if it were a permanent magnet. Electromagnets are useful because they can be activated and deactivated by closing and opening a circuit.To learn more about electromagnet, refer to
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A sports car accelerates from rest to 95km/h in 6.2s.what is the average acceleration in m/s
Answer:
4.3 m/s^2
Explanation:
V=0
\(V_{f}\) = 26.41 m/s
t= 6.2 s
a= ?
\(V_{f}\) = \(V_{i}\) + at
26.41 m/s = 0+a (6.2s)
a= \(4.3 m/s^{2}\)
A grasshopper is an insect that goes through three distinct stages during its life cycle. The life cycle of a grasshopper is an example of which one of the following?
A. Quick metamorphosis
B. Slow metamorphosis
C. Complete metamorphosis
D. Incomplete metamorphosis
Answer:
B is the answer bc it's no way c an a and d can maybe c
What is the difference between elastic PE and gravitational PE?
1. Two charges are separated by a distance of 1 cm. One charge has a value of 7 micro Coulombs. The other charge has a value of 10 micro Coulombs. What is the force between them, in pounds. Make sure to include the sign of the force which will be positive if the charges repel each other and negative if they attract each other.
2. 12 gauge copper wire is normally used in house wiring. When aluminum wire is used one needs to use a smaller gauge size to obtain the same resistance, 40 ft of 12 gauge copper wire was calculated. What would the resistance be if 10 gauge aluminum wire were used?
3. A 12 V automobile battery can supply 51 amps for one hour and cost $194. What is the cost of this electricity in cents per kWh?
4. Most of the body's resistance is in its skin. When wet, salts go into ion form, and the resistance is lowered. Thus, the resistance of the skin can go from 100,000 ohms when dry to 300 ohms when wet. What is the current that would be carried through the body, in milliAmperes, if you touched a 240 V power line while dry? Currents over 10 mA are almost always deadly.
1. The force between the two charges is 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.
2. The resistance of 10 gauge aluminum wire over a 40 ft distance would be 0.506 ohms.
3. The cost of electricity from the automobile battery is 38.6 cents per kWh.
4. The current that would be carried through the body is 0.8 mA if dry.
1. The force between two point charges can be calculated using Coulomb's law, which states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Using the values given, the force can be calculated as F = (k * q1 * q2) / r², where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. Plugging in the values, the force can be calculated as 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.
2. The resistance of a wire is determined by its length, cross-sectional area, and resistivity. The resistivity of aluminum is higher than that of copper, so a larger cross-sectional area is required to achieve the same resistance. Using the gauge size conversion chart, 10 gauge aluminum wire has a cross-sectional area of 5.26 mm², which is approximately 83% of the cross-sectional area of 12 gauge copper wire.
Thus, the resistance of 10 gauge aluminum wire over a 40 ft distance can be calculated as R = (rho * L) / A, where rho is the resistivity of aluminum, L is the length, and A is the cross-sectional area. Plugging in the values, the resistance can be calculated as 0.506 ohms.
3. To calculate the cost of electricity per kWh, the total cost and the total amount of energy supplied must be known. Since the battery supplies 12 V and 51 A for one hour, the total energy supplied can be calculated as E = V * I * t, where V is the voltage, I is the current, and t is the time.
Plugging in the values, the total energy supplied can be calculated as 612 watt-hours (Wh). Since one kWh is equal to 1000 Wh, the total energy supplied can be converted to 0.612 kWh. Dividing the total cost by the total energy supplied gives the cost per kWh, which is 38.6 cents.
4. The current through the body can be calculated using Ohm's law, which states that current is equal to voltage divided by resistance. Using the values given, the resistance can be either 100,000 ohms or 300 ohms depending on whether the skin is dry or wet.
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An object is moving with an initial velocity of 5.5m/s.It is then subject to a constant acceleration of 2.5 m/s for 11s. How far will it have traveled during the time of its acceleration?
The distance traveled by the object during the time of acceleration is 211.75 m.
What is distance?Distance can be defined as the total lenght between two points.
T o calculate the distance traveled by the object during the time of acceleration, we use the formula below.
Formula:
s = ut+at²/2............ Equation 1Where:
s = Distanceu = Initial velocitya = Accelerationt = TimeFrom the question,
Given:
u = 5.5 m/st = 11 sa = 2.5 m/s²Substitute these values into equation 1
s = (5.5×11)+(2.5×11²)/2s = 60.5+151.25s = 211.75 mHence, the distance traveled by the object is 211.75 m.
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The diagram shows four pairs of large parallel conducting plates. The value of the electric potential is given for each plate. Rank the pairs according to the magnitude of the electric field between the plates, least to greatest
The magnitude of the electric field between the plates, from least to greatest is shown by option C.
What is the ranking?We know that when two plates that are conducting are arranged in such a way that a gap is left in between them, the we have a capacitor. The electric potential of the capacitor would have a lot to do with the distance of the separation of the plates.
As such, the closer the plates are together, the greater the electric potential of the plates and the more the plates are apart, the lesser the electric potential of the plates.
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True or false energy must be added to the gas element in order for light to be emitted .
A. True
B. False
Answer:
b is the answer of this qs
Suggest two reasons why using a graph to find the refractive index is a better method than simply calculating it using a pair of angles from the table.
Two reasons why using a graph to find the refractive index is a better method than simply calculating it using a pair of angles from the table are:
i) Graphs provide a better visual representation of the data: When calculating the refractive index using a pair of angles from a table, it can be difficult to visualize how the data points relate to each other.
ii) Calculating refractive index using a graph is more accurate: Calculating the refractive index using a pair of angles from a table can be prone to errors, particularly if the table is not accurate or the measurements are imprecise.
When it comes to finding the refractive index, using a graph is often a better method than simply calculating it using a pair of angles from the table.
Here are two reasons why:
Graphs provide a better visual representation of the data:When calculating the refractive index using a pair of angles from a table, it can be difficult to visualize how the data points relate to each other. By contrast, a graph provides a clear and easy-to-read visual representation of the data, making it easier to identify patterns and trends. Graphs are also useful for identifying outliers or errors in the data that might not be immediately apparent when using a table.
Calculating refractive index using a graph is more accurate: Calculating the refractive index using a pair of angles from a table can be prone to errors, particularly if the table is not accurate or the measurements are imprecise. Using a graph, on the other hand, allows for a more precise calculation of the refractive index. This is because the graph allows you to take into account any minor variations in the measurements, which might be missed when using a table.
Overall, using a graph is a more accurate and reliable method for finding the refractive index than simply using a table. It allows for a better visual representation of the data and provides a more accurate calculation of the refractive index.
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Explain your answer. What parts of your hypothesis were strong correct? What parts were weak?
Use the data from the table and the Punnett Square in your explanation.
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So better check this question!
A spring whose spring constant is 850 N/m is compressed 0.40 m. What is the maximum speed it can give to a 500. g ball?
4. If the spring in #3 were compressed twice as much, how many times greater would the velocity of the ball be?
Answer:
part 1. 1/2kx^2 = 1/2mv^2
1/2(850)(0.40)^2 = 1/2(0.5)v^2
68 = 0.25v^2
272 = v^2
v = 15.49m/s
Explanation:
part 1. the max potential energy would be equal to the maximum kinetic energy assuming no energy is lost, because of this, we can set the spring potential energy to the kinetic energy. from there plug in all the variables sorry all known variables and solve for v. make sure to plug in 0.5 for mass so that it is in kilograms.
often occurs at subduction zones. Check all that apply:
O convergence
trench
upward convection currents in the mantle below it
Orifts
most volcanoes...Pacific Ring of Fire
continental plate subducts because it's less dense
Answer:
convergence and trench
Explanation:
LOL
a sleepy student drops a calculator out of a window. we can ignore air resistance. what is the velocity of the calculator after falling for 1.8s?
Answer: -18
Explanation:
Answer:
its -18
Explanation:
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