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matrenka [14]
3 years ago
15

student built a simple heat engine that could lift masses. If the heat engine takes in 300 J of heat and loses 50 J to the envir

onment, how much height will it lift a 1.5 kg mass?
Physics
1 answer:
k0ka [10]3 years ago
4 0

Answer:

So it will lift the mass by h = 17 m

Explanation:

As per energy conservation we know that

Q_1 = W + Q_2

here we know that

Q_1 = 300 J

Q_2 = 50 J

now we have

W = 300 - 50

W = 250 J

so work done by the engine is 250 J

now we have

W = mgh

250 = 1.5 \times 9.81 \times h

h = 17 m

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Determine the volume of a cube whose width is 32.1 cm. Round your answer to the correct number significant figures.
Romashka [77]

The volume of a cube is given by:

V=L^3

where L is the measure of each side (which corresponds to its width, since all sides of a cube are equal).

Plugging the data of the problem, L=32.1 cm, inside the equation, we find the volume:

V=L^3 =(32.1 cm)^3 =33076.2 cm^3

4 0
4 years ago
A rock at rest has weight 138 n. what is the weight of the rock when it is accelerating downward at 3.5 m/s2?
tamaranim1 [39]
<span>The weight of the rock is 138n when accelerating downward at 3.5m/ s2.Weight is a measurement of the force of gravity on an objects mass. Therefore as long as the force of gravity is the same and the mass of the rock is the same the weight should also be the same.</span>
7 0
3 years ago
A mountain climber stands at the top of a 50.0 m cliff hanging over a calm pool of water. The climber throws two stones vertical
Sonbull [250]

Answer:

a. t=2.997\ s is the time after which the two stone hit the water surface.

b. u_2=1.998\ m.s^{-1}

c. v_1=31.3688\ m.s^{-1}   &  v_2=31.3686\ m.s^{-1}

Explanation:

Given:

  • height of the cliff, h=50\ m
  • time gap between the projection of stones, \Delta t=1\ s
  • initial velocity of first stone, u_1=+2\ m.s^{-1}

Here positive sign means that the stone is thrown vertically downward in the direction of gravity.

a.

Using equation of motion:

h=u_1.t+\frac{1}{2} g.t^2

50=2t+4.9t^2

t=2.997\ s is the time after which the two stone hit the water surface.

b.

using the equation of motion for second stone:

h=u_2.t+\frac{1}{2} g.t^2

50=u_2\times 2.997+4.9\times 2.997^2

u_2=1.998\ m.s^{-1}

c.

Final velocity of the stone at the instant of touching the water surface:

v_1^2=u_1^2+2.g.h

v_1^2=2^2+2\times 9.8\times50

v_1=31.3688\ m.s^{-1}

&

v_2^2=u_2^2+2g.h

v_2^2=1.998^2+2\times 9.8\times 50

v_2=31.3686\ m.s^{-1}

6 0
4 years ago
if the magnitudes of the forces vary with time as F1=Ct and F = 2Ct, where C equals to 7.5 N/s and t is time, find the time t0 a
Degger [83]

Answer:

The tension in the string is equal to Ct

And the time t0 when the rension in the string is 27N is 3.6s.

Explanation:

An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.

So T = Ct

When T = 27N then t = T/C = 27/7.5 = 3.6s

4 0
3 years ago
A leopard chasing a deer reaches a constant speed of 75m / s. In How long does 1200m travel?
xxTIMURxx [149]

Answer:

16 seconds

Explanation:

To do this you would set up the equation 75x=1200. Then you would divide 75 from both sides so you get x=16 or it takes 12 seconds for it to travel 1200 meters

6 0
4 years ago
Read 2 more answers
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