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3 years ago

How does a change in thermal energy cause matter to change from one state to another?

1 answer:

4 0

Thermal energy gives the particles of the substance kinetic energy because temperature is an average measure of kinetic enegy of the particle. If we give them thermal energy the particle will move faster, gaining enough energy to escape and become free. For example, from solid to liquid, the particles would espace their fixed position and be free to move as a liquid.

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A boxed 12.0 computer monitor is dragged by friction 7.50 up along the moving surface of a conveyor belt inclined at an angle of

sasho [114]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below is the solution:

W done by Normal = 0. (make the incline flat, Normal force goes directly up: no work done)
W done by gravity = w*displacement = (11kg*9.8) * 7.5sin(35) = -463J 
W done by friction is the opposite of the work done by weight because the object is not moving. Therefore W done by friction = 463J

8 0

4 years ago

Jane placed two bar magnets on her desk next to each other with like poles facing. She then sprinkled some iron filings around t

viktelen [127]

Answer:

Explanation:

3 0

4 years ago

Read 2 more answers

A car of mass 1000 kg travelling at a velocity of 25 m/s collides with another car of mass 1500kg which is at rest. The two cars

Svetach [21]

Answer:

The velocity of the two cars is 10 m/s after the collision.

Explanation:

Law Of Conservation Of Linear Momentum

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is

P=m.v

If we have a system of bodies, then the total momentum is the sum of them all

$P=m_1v_1+m_2v_2+...+m_nv_n$

If some collision occurs, the velocities change to v' and the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

In a system of two masses, the law of conservation of linear momentum takes the form:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The car of mass m1=1000 Kg travels at v1=25 m/s and collides with another car of m2=1500 Kg which is at rest (v2=0).

Knowing both cars stick and move together after the collision, their velocity is found solving for v':

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

$\displaystyle v'=\frac{1000*25+1500*0}{1000+1500}$

$\displaystyle v'=\frac{25000}{2500}$

v' = 10 m/s

The velocity of the two cars is 10 m/s after the collision.

4 0

3 years ago

Please helppp ! Is my choice right ?

grin007 [14]

Yes you are right because if both of the photons are the same nothing is really happening thereafter c is the correct answer.

6 0

3 years ago

If a projectile is fired with an initial velocity of v0 meters per second at an angle α above the horizontal and air resistance

lawyer [7]

Answer:

a) The bullet hits the ground 51.02 s after it was fired.

b) 22,092.3 units

c) 3188.8 units

Explanation:

a) Assuming that the level the bullet was fired from is the ground level.

The bullet hits the ground when y = 0

y = (v₀ sin(α))t − (1/2) gt²

v₀ = 500 m/s

α = 30°

y = 0

0 = (500 sin 30) t - 0.5(9.8)t²

4.9t² - 250t = 0

t(4.9t - 250) = 0

t = 0 s or (4.9t - 250) = 0

The t = 0 s indicates that the bullet was indeed fired from the ground level.

The time it eventually hits the ground back

4.9t = 250

t = 51.02 s

The bullet hits the ground 51.02 s after it was fired.

b) The distance from the firing point that the bullet lands.

x = (v₀ cos(α))t

At this horizontal distance, t = 51.02 s

Substituting the parameters

x = (500 cos 30°) × 51.02

x = 22,092.3 units

c) Maximum height attained by the bullet

Maximum height is given by

H = (u² sin² α)/2g

H = (500² sin² 30°)/(2×9.8)

H = 3188.8 units

Hope this Helps!!!

6 0

4 years ago