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galina1969 [7]
4 years ago
12

8. A student does 1,000 J of work when she moves to her dormitory. Her internal energy is decreased by 3,000 J. Determine the he

at during this process. Does she gain or lose her heat
Physics
1 answer:
Lyrx [107]4 years ago
6 0

Answer:

The heat loss during the process = -4000 J

Explanation:

Work done by the student (W) = - 1000 J

Negative sign on the system is due to work done on the system.

Decrease in internal energy (U) = - 3000 J

We know that heat transfer in the system is given by (Q) = U + W

⇒ Q = - 1000 - 3000

⇒ Q = - 4000 J

This is the value of heat transfer during the process And negative sign indicates that heat loss during the process.

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You exert a force of 25 newtons while you move a rock 15 meters. How much work did you perform?
leonid [27]
Work Done = Force x Distance Moved
Work Done = 25 x 15 = 375 Joules
4 0
3 years ago
Calculate the gravitational potential energy of a 1200 kg car at the top of a hill that is 42 m high.
GalinKa [24]
It IS <span>PE = (1200 kg)(9.8 m/s²)(42 m) = 493,920 J </span>
6 0
3 years ago
A beaker of negligible heat capacity contains 456 g of ice at -25.0°C. A lab technician begins to supply heat to the container a
Scorpion4ik [409]

Answer:

176 min

Explanation:

456 g = .456 kg

Specific heat of ice s = 2093 J kg⁻¹

Heat required to raise the temperature by 25 degree

= mass x specific heat x rise in temperature.

= .456 x 2093 x 25

=23860 J

Heat required to melt the ice to make water at zero degree

= mass x latent heat

= .456 x 334 x 10³

=152304 J

Total heat required = 152304 + 23860 = 176164 J .

Time Required = Heat required / rate of supply of heat

= 176164 / 1000

176.16 min

4 0
3 years ago
What is the z coordinate for the above vector?
Hitman42 [59]
120 m would be he answer
4 0
3 years ago
An 18g bullet is shot vertically into a 10kg block. The block lifts upward 9mm. The bullet penetrates the block in a time interv
stealth61 [152]

Answer:

The initial kinetic energy of the bullet is closest to 491.87 J

Explanation:

Given;

mass of bullet, m₁ = 18g = 0.018kg

mass of block, m₂ = 10kg

height moved by the block, h = 9 mm = 0.009 m

time taken for the bullet to travel through the block, t = 0.001s

let the initial velocity of the bullet = v₁

let the final velocity of the bullet = v₂

Apply the principle of conservation of linear momentum;

initial momentum = final momentum

0.018v₁ = v₂(0.018 + 10)

0.018v₁ = 10.018v₂ -----equation (1)

Apply the law of conservation of energy when the bullet lifts the block through 9mm

mgh = ¹/₂mv₂²

gh = ¹/₂v₂²

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.009)

v₂ = 0.42 m/s

Substitute in v₂ in equation 1, to determine the initial velocity of the bullet;

0.018v₁ = 10.018v₂

0.018v₁  =  10.018(0.42)

0.018v₁  = 4.208

v₁ = 4.208 / 0.018

v₁ = 233.78 m/s

Now, determine the initial kinetic energy of the bullet;

K.E₁ = ¹/₂m₁v₁²

K.E₁ = ¹/₂(0.018)(233.78)²

K.E₁ = 491.87 J

Therefore, the initial kinetic energy of the bullet is closest to 491.87 J

6 0
3 years ago
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