Ask question

Ask question

All categories

3 years ago

12

Question 1 of 5 Which term describes a measurement of the movement of atoms or molecules within an object? O A. Temperature O B.

Heat O C. Pressure O D. Density

1 answer:

Fittoniya [83]3 years ago

7 0

Answer:

Temperature

Explanation:

You might be interested in

When resistance force is increased on a lever which of the following happens to the effort force?

Sauron [17]

When resistance force on a lever increases, nothing happens automatically.
But if you want to keep lifting the load, then YOU must increase the force of
your effort in order to make it happen.

8 0

3 years ago

Would you classify petroleum as a renewable or nonrenewable resource? Justify your answer in two or more complete sentences.

olchik [2.2K]

Answer:

Petroleum is a nonrenewable resource. This being because it is a natural gas that take millions of years for it to reform which means each part of it is a once in a lifetime use.

3 0

3 years ago

An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit

lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is $1.127\times 10^7m/s$

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

$m_1v_1=m_2v_2$

where,

$m_1\text{ and }v_1$ = Initial mass and velocity

$m_2\text{ and }v_2$ = Final mass and velocity

We are given:

$m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s$

Putting values in above equation, we get:

$238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s$

Hence, the velocity of released alpha particle is $1.127\times 10^7m/s$

4 0

3 years ago

A track star in the broad jump goes into the jump at 12 m/s and launches himself at 20° above the horizontal. How long is he in

emmainna [20.7K]

Explanation:

It is given that,

Initial speed of the broad jump, u = 12 m/s

It is launched at an angle of 20 degrees above the horizontal. Let t is the time for which the track star i in the air before returning to Earth. The motion of the track star in the broad jump can be treat as the projectile motion. The time of flight of the projectile is given by :

$t=\dfrac{2u\ sin\theta}{g}$

Putting all the values in above equation as :

$t=\dfrac{2\times 12\times \ sin(20)}{9.8}$

t = 0.837 seconds

So, the time for which the track star is in air is 0.837 seconds. Hence, this is the required solution.

3 0

3 years ago

Gwar [14]

<u>Answer:</u>

For 1: The correct option is Option C.

For 3: The final velocity of the opponent is 1m/s

<u>Explanation: </u>

During collision, the energy and momentum remains conserved. The equation for the conservation of momentum follows:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...(1)

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of first object

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of second object

<u>For 1:</u>

We are Given:

$m_1=150g=0.15kg\\u_1=?m/s\\v_1=0.85m/s\\m_2=3500g=3.5kg\\u_2=0m/s\\v_2=0.85m/s$

Putting values in equation 1, we get:

$(0.15\times u_1)+(3.5\times 0)=(3.5+0.15)\times 0.85\\\\u_1=20.683\approx 21m/s$

Hence, the correct answer is Option C.

<u>For 2: </u>

Impulse is defined as the product of force applied on an object and time taken by the object.

Mathematically,

$J=F\times t$

where,

F = force applied on the object

t = time taken

J = impulse on that object

Impulse depends only on the force and time taken by the object and not dependent on the surface which is stopping the object.

Hence, the impulse remains the same.

<u>For 3:</u>

Let the speed in right direction be positive and left direction be negative.

We are Given:

$m_1=240kg\\u_1=0m/s\\v_1=-1m/s\\m_2=80kg\\u_2=-2m/s\\v_2=?m/s$

Putting values in equation 1, we get:

$(240\times 0)+(80\times (-2))=(240\times (-1))+(80\times v_2)\\\\v_2=1m/s$

Hence, the final velocity of the opponent is 1m/s and has moved backwards to its direction of the initial velocity.

4 0

3 years ago