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3 years ago

If 2 objects had the same momentum, what must be true about the mass of the object that traveled the fastest?

Physics

1 answer:

julsineya [31]3 years ago

4 0

Yes, the above-given statement is true

<u>Explanation:</u>

The product of the mass x the velocity will be the same for both. Momentum is the action of a body with a particular mass through space and there is the conservation of momentum.
Momentum is described as the mass of the object multiplied by its velocity.
<u>Momentum (p) = Mass (M) * Velocity (v)</u>
Therefore for two objects with many masses to have a similar momentum, then the lighter one has to be moving quicker than the heavier object.

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A(n) 1.3 kg mass sliding on a frictionless surface has a velocity of 7.1 m/s east when it undergoes a one-dimensional elastic co

Oxana [17]

Answer: 2.12 kg

Explanation:

Since the 1.3 kg object moves to the west after the collision, the other object will move to the east after the collision.

In an elastic collision, the relative velocity after the collision is the opposite of the relative velocity before the collision. Since the 1.3 kg object’s velocity before the collision is 6.7 m/s greater than the other object, after the collision, its velocity will be 6.7 m/s less than the other object. To determine the other object’s velocity, use the following equation.

v = 1.7 – 7.1 = -5.4 m/s

The negative sign means it is moving eastward. Let’s use this number is a momentum equation to determine its mass.

Initial momentum = 1.3 * 7.1 = 9.23 east

For the 1.3 object, final momentum = 1.3 * 1.7 = 2.21 west

To determine the final momentum of the other object, add these two numbers.

Final momentum = 11.44 east

To determine its mass, use the following equation.

m * 5.4 = 11.44

m = 11.44 ÷ 5.4 = 2.12 kg

To make sure that kinetic energy is conserved, let’s round this number to 2 kg and determine the final kinetic energies.

For the 1.3 object, KE = 1/2 * 1/3* 1.7^2 = 0.48

For the 2 kg object, KE = 1/2* 2 * 5.4^2 = 29.64

Total final KE = 29.64

Initial KE = 0.5* 1.3 * 7.1^2 = 32.77

Since I rounded the mass up to 2kg, this proves that kinetic energy is conserved and the mass is correct!

3 0

3 years ago

Which of the following are early clues that signal an earthquake may occur? Select all that apply. Changes in magnetic propertie

enot [183]

The following answers apply;

Changes in magnetic properties of rock
Decrease in well water levels
Increases in radon gas in groundwater
Foreshocks

These other choices may be good indicators of an imminent volcanic eruption;

Movement of magma

Increase in sulfur dioxide and carbon dioxide ground emissions

6 0

3 years ago

Read 2 more answers

In Physics, work depends on two factors. What are those two<br> factors?

Aloiza [94]

Answer:

Force and displacement.

Explanation:

Work done is positive when we push table and it move in the direction of applied force.

5 0

3 years ago

A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a

wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

$\frac{V}{T}=const.$

Which can be also re-written as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where

$V_1, V_2$ are the initial and final volumes of the gas

$T_1, T_2$ are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

$V_1 = 7.00\cdot 10^2 mL = 700 mL$ is the initial volume

$T_1=20.0^{\circ}C+273=293 K$ is the initial absolute temperature

$V_2$ is the final volume

$T_2 = 1.00\cdot 10^2 K = 100 K$ is the final temperature

Solving for $V_2$ ,

$V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL$

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

6 0

3 years ago

What quantity of heat must be removed from 20g<br>of water at 0°C to change it to ice at 0°C?

seraphim [82]

The quantity of heat must be removed is 1600 cal or 1,6 kcal.

<h3>Explanation : </h3>

From the question we will know if the condition of ice is at the latent point. So, the heat level not affect the temperature, but it can change the object existence. So, for the formula we can use.

$\boxed {\bold {Q = m \times L}}$

If :

Q = heat of latent (cal or J )
m = mass of the thing (g or kg)
L = latent coefficient (cal/g or J/kg)

<h3>Steps : </h3>

If :

m = mass of water = 20 g => its easier if we use kal/g°C
L = latent coefficient = 80 cal/g

Q = ... ?

Answer :

$Q = m \times L \\ Q = 20 \times 80 = 1600 \: cal$

So, the quantity of heat must be removed is 1600 cal or 1,6 kcal.

<u>Subject : Physics </u>

<u>Subject : Physics Keyword : Heat of latent</u>

4 0

3 years ago