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nasty-shy [4]
3 years ago
15

Describe what happens when heat is removed from a system

Physics
1 answer:
OleMash [197]3 years ago
8 0
This happens<span> at the boiling </span>temperature<span> of every substance that can vaporize. At the boiling </span>temperature<span>, adding </span>heat<span> energy converts the liquid into a gas WITHOUT RAISING THE </span>TEMPERATURE<span>. Adding </span>heat<span> to a boiling liquid is an important exception to general rule that more </span>heat<span> makes a higher </span>temperature<span>.</span>
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A boat is traveling with a velocity of 18 meters/second relative to water and the river is flowing at a velocity of 2.5 meters/s
ddd [48]
Vt = Vboat - Vriver 
Vt = 18 - 2.5 = 15.5 m/s 

If the boat's direction is the same as the water, you sum the velocities of the river and the boat . 
8 0
2 years ago
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In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves wit
Minchanka [31]

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}

Putting, t = 1 s  and x = 4 m in above equation, we get :

4 = \dfrac{a(1)^2}{2}\\\\a = 8 \  m/s^2

Therefore, the acceleration of the cart is 8 m/s².

5 0
2 years ago
The distance between two charges a and b is r, and the force between them is F. What is the force between them if the distancbet
Nadusha1986 [10]
See coulomb's law. Force is inversely proportional to the distance squared. So if you multiply r by 2, the force is multiplied by (½)² = ¼.

a. F/4
7 0
3 years ago
(e)
Hatshy [7]

Answer:

≈933.3kg/m^3

Explanation:

Density=Mass/Volume

11200kg/12.0= 933.3333kg/m^3

5 0
3 years ago
The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer
makkiz [27]

Answer:

A \to rad/s

B  \to rad/s^3

Explanation:

\omega_z(t)=A + Bt^2

Required

The units of A and B

From the question, we understand that:

\omega_z(t) \to rad/s

This implies that each of A and Bt^2 will have the same unit as \omega_z(t)

So, we have:

A \to rad/s

Bt^2 \to rad/s

The unit of t is (s); So, the expression becomes

B * s^2 \to rad/s

Divide both sides by s^2

B  \to \frac{rad/s}{s^2}

B  \to rad/s^3

5 0
2 years ago
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