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3 years ago

What happens when heat is removed from a liquid and it cools?

Physics

2 answers:

saul85 [17]3 years ago

5 0

The energy content is decreased and so is the particle speed.

nikitadnepr [17]3 years ago

4 0

Than it hardens and turns into a solid

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What part of the water is affected by a wave in deep water

krok68 [10]

Wave base is the most affected by a wave in deep water. :]

5 0

3 years ago

a box is at rest on a ground. An unbalance force acts on the box causes it to start. The box moves 10 meters in 2 seconds and th

nordsb [41]

The box stopped moving because there was a negative acceleration in the x direction caused by friction.
To find the average speed, simply divide the displacement by time: 10 / 2 = 5 m/s

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2 years ago

Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.

Anni [7]

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)= $k q1q2/r^2$

$k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m$

So,

$F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}$

$F= 0.6 N$

5 0

3 years ago

An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be

lord [1]

Answer:

the quarterback will be moving back at speed of 0.080625 m/s

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0

3 years ago

A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the

ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

$\frac {Vf^{2}-Vo^2}{2.a} =X$