As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. increases proportionally D. decreases proportionally

Answers

Answer 1

Answer: the answer is c. as an object is in motion speeds up or “becomes heavier”, it’s kinetic energy increases proportionally: double the velocity and you quadruple the kinetic energy. this is why a tiny bullet traveling at high speed does so much more damage than a huge truck bumping into something at 1 mph. so the answer is c

Answer 2

Answer:

Answer:

Explanation:

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In the calorimetry experiment which energy will be calculated during the heat exchange if water is used?

Answers

From the coffee cup to the thermometer

The assumption behind the science of calorimetry is that the energy gained or lost by the water is equal to the energy lost or gained by the object under study. So if an attempt is being made to determine the specific heat of fusion of ice using a coffee cup calorimeter, then the assumption is that the energy gained by the ice when melting is equal to the energy lost by the surrounding water. It is assumed that there is a heat exchange between the iceand the water in the cup and that no other objects are involved in the heat exchanged. This statement could be placed in equation form as

Qice = - Qsurroundings = -Qcalorimeter

The role of the Styrofoam in a coffee cup calorimeter is that it reduces the amount of heat exchange between the water in the coffee cup and the surrounding air. The value of a lid on the coffee cup is that it also reduces the amount of heat exchange between the water and the surrounding air. The more that these other heat exchanges are reduced, the more true that the above mathematical equation will be. Any error analysis of a calorimetry experiment must take into consideration the flow of heat from system to calorimeter to other parts of the surroundings. And any design of a calorimeter experiment must give attention to reducing the exchanges of heat between the calorimeter contents and the surroundings.

The energy calculated while dealing with the calorimeter experiment are the latent heat of vaporization, latent heat of fusion and the heat required to change the temperature of the substances.

Further Explanation:

The calorimeter works on the principle of conservation of energy. The amount of heat given by one part of the system is equal to the amount of heat gained by another part provided that the calorimeter does not loss any heat to the environment.

Consider that ice is mixed with water at some temperature. Then the water being at higher temperature losses heat to the ice at lower temperature. The ice gains the heat from the water and the system reaches an equilibrium at which the solution of ice and water has the same amount of energy at a particular temperature.

The different types of energies dealt with in the calorimetry experiment are as follows:

Latent heat of fusion:

The amount of energy required by a body when it is melted from its frozen state or freezes from its melted state is termed as the latent heat of fusion.

For example:

The small amount of ice is mixed with water in a calorimeter. Here, the ice requires the latent heat of fusion that leads to the melting of ice and converts it into water.

Latent heat of vaporization:

The amount of heat required to convert one gram of liquid to vapor without raising its temperature is known as latent heat of vaporization.

For example:

The water is boiling at in a calorimeter. Here, the water requires latent heat of vaporization which leads to the vaporization of water and convert it into vapors.

Thus, the latent heat of fusion, latent heat of vaporization and the heat required to change the temperature of the substance are the energies measured with the calorimeter.

Learn more:

1. Transnational kinetic energy brainly.com/question/9078768.

2. Expansion of gas brainly.com/question/9979757.

3. Conservation of momentum brainly.com/question/9484203.

Answer Details:

Grade: College

Subject: Physics

Chapter: Heat and Energy

Keywords:

Heat, energy, calorimeter, latent heat, vaporization, fusion, experiment, temperature, melting, boiling, liquid, vapor, evaporation, condensation, freeze.

Two charged metallic spheres of radii, R = 10 cms and R2 = 20 cms are touching each other. If the charge on each sphere is +100 nC, what is the electric potential energy between the two charged spheres?

Answers

Answer:

$200* 10^(-6)j$

Explanation:

We have given the radius of first sphere is 10 cm and radius of second sphere is 20 cm

So the potential of first sphere will be greater than the potential of the second sphere, so charge will flow from first sphere to second sphere

Let q charge is flow from first sphere to second sphere and then potential become same

So $V=(K(100-q))/(r_1)=(K* 100)/(r_2)$

200-100=2q+q

$q=(100)/(3)=33.33nC$

So $V=(K(100-q))/(r_1)=(9* 10^(9)* (100-33.33)* 10^(-9))/(10* 10^(-2))=6003V$

We know that potential energy U=qV $=33.33* 10^(-9)* 6003=200* 10^(-6)j$

Answer:

The electric potential energy between the two charged spheres is $199.9*10^(-6)\ J$

Explanation:

Given that,

Radius of first sphere $R_(1)=10\ cm$

Radius of second sphere $R_(2)=10\ cm$

Charge Q= 100 nC

We know charge flows through higher potential to lower potential.

Using formula of potential

$V=(k(Q-q))/(R_(1))$ ...(I)

$V=(k(Q+q))/(R_(2))$ ...(II)

From equation (I) and (II)

$(k(Q-q))/(R_(1))=(k(Q+q))/(R_(2))$

Put the value into the formula

$((100-q))/(10*10^(2))=((100+q))/(20*10^(-2))$

$(100-q)*20*10^(-2)=(100+q)*10*10^(-2)$

$q=(1000)/(30)$

$q=(100)/(3)$

$q=33.33\ nC$

So, the potential at R₁ and R₂

Using formula of potential

$V=(k(Q-q))/(R_(1))$

Put the value into the formula

$V=(9*10^(9)(100-33.33)*10^(-9))/(10*10^(-2))$

$V=6000.3\ Volt$

We need to calculate the electric potential energy between the two charged spheres

Using formula of the electric potential energy

$U=qV$

$U=33.33*10^(-9)*6.0003*10^(3)$

$U=199.9*10^(-6)\ J$

Hence, The electric potential energy between the two charged spheres is $199.9*10^(-6)\ J$

A toy car that is 0.12 m long is used to model the actions of an actual car thatis 6 m long. Which ratio shows the relationship between the sizes of the
model and the actual car?
A: 5:1
B: 1:50
C: 50:1
D: 1:5

Answers

Answer:

it is B 1:50

Explanation:

just did it on apex

Final answer:

The ratio that shows the relationship between the sizes of the model car and the actual car is 1:50. This is because the actual car is 50 times longer than the model car.

Explanation:

The relationship between the sizes of the model car and the actual car is represented by a ratio. To find this ratio, we can divide the length of the actual car by the length of the model car. So, 6 m / 0.12 m = 50. This means that the actual car is 50 times longer than the model car, or in other words, the model car is 1/50th the size of the actual car. Therefore, the correct ratio is 1:50.

Learn more about Ratios here:

brainly.com/question/32531170

#SPJ3

Twopucksofequalmasscollideonafrictionlesssurface,asillustratedinthefigure.Immediatelyafterthe collision, the speed of the black puck is 1.5 m/s. What is the speed of the white puck immediately after the collision?