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3 years ago

What do we call the energy that is transferred to

Physics

1 answer:

mojhsa [17]3 years ago

5 0

Answer:

Latent heat

Latent heat is the heat needed to change a state without a change in temperature.

Explanation:

Hope this helps :D

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A hydraulic lift is made by sealing an ideal fluid inside a container with an input piston of cross-sectional area 0.004 m2 , an

kvv77 [185]

Answer:

Maximum weight that can be lifted = 18,000 N

Explanation:

Given:

Cross-sectional area of input (A1) = 0.004 m²

Cross-sectional area of the output (A2) = 1.2 m ²

Force (F) = 60 N

Computation:

Pressure on input piston (P1) = F / A1

Assume,

Maximum weight lifted by piston = W

Pressure on output piston (P2) = W / A2

We, know that

P1 = P2

[F / A1] = [W / A2]

[60 / 0.004] = [W / 1.2]

150,00 = W / 1.2

Weight = 18,000 N

Maximum weight that can be lifted = 18,000 N

3 0

3 years ago

a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what

Andrews [41]

Answer:

v=9.6 km/s

Explanation:

Given that

The mass of the car = m

The mass of the truck = 4 m

The velocity of the truck ,u= 12 km/s

The final velocity when they stick = v

If there is no any external force on the system then the total linear momentum of the system will be conserve.

Pi = Pf

m x 0 + 4 m x 12 = (m + 4 m) x v

0 + 48 m = 5 m v

5 v = 48

$v=\dfrac{48}{5}\ km/s$

v=9.6 km/s

Therefore the final velocity will be 9.6 km/s.

3 0

3 years ago

Four vehicles approach an intersection with a 4 way stop at the same time. Car B is ahead of Car A and both are in the same lane

garik1379 [7]

Answer: car D

Explanation:

4 0

3 years ago

A 95 kg clock initially at rest on a horizontal floor requires a 650 n horizontal force to set it in motion. after the clock is

Tamiku [17]

The μs between the clock and floor is 650(M*g) and the μk between the clock and the floor is 560(M*g)

3 0

3 years ago

Read 2 more answers

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago