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jek_recluse [69]
2 years ago
11

It takes a crane 59s to lift a flagstone using 342 W of power. How much work is done on the flagstone? ​

Physics
1 answer:
zepelin [54]2 years ago
5 0

Answer: The work done on the flagstone is 20178 J

Explanation:

Power is the rate at which work is done . It is equal to the amount of work done divided by the time it takes to do the work.

Work=Power\times time

Given : work = ?

Power = 342 W = 342J/s

Time = 59 s

Work=342J/s\times 59s=20178J

Thus the work done on the flagstone is 20178 J

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erastova [34]

Answer:

m

Explanation:

3 0
2 years ago
A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin
ollegr [7]
Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
pV=k
which can also be rewritten as
p_1 V_1 = p_2 V_2 (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, p_1 = 1.5 atm, V_1 =6.5 L and V_2 =13.0 L, so the final pressure of the gas can be found by re-arranging eq.(1):
p_2 = p_1  \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm

Therefore the correct answer is
<span>1. 0.75 atm</span> 
8 0
3 years ago
An aluminum cup of 150 cm3 capacity is completely filled with glycerin at 23°C. How much glycerin will spill out of the cup if t
AVprozaik [17]

Answer:

1.19cm^3 of glycerine

Explanation:

Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:

Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum

Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature

coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1

Change in temperature = 41-23 = 18oC

Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3

3 0
2 years ago
A woman on a bridge 82.2 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an a
solong [7]

Answer:

0.71 m/s

Explanation:

We find the time it takes the stone to hit the water.

Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.

So, substituting the values of the variables into the equation, we have

y = ut - 1/2gt²

82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²

82.2 m = 0 + (4.9 m/s²)t²

82.2 m =  (4.9 m/s²)t²

t²  = 82.2 m/4.9 m/s²

t² = 16.78 s²

t = √16.78 s²

t = 4.1 s

This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.

Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s

5 0
3 years ago
The trough of the sine curve used to represent a sound wave corresponds to
iren [92.7K]

Answer:

The correct answer is a rarefaction.

Explanation:

Sound waves are longitudinal waves that propagate in a medium, such as air. As the vibration continues, a series of successive condensations and rarefactions form and propagate from it. The pattern created in the air is something like a sinusoidal curve to represent a sound wave.

There are peaks in the sine wave at the points where the sound wave has condensations and valleys where it has rarefactions.

Have a nice day!

4 0
3 years ago
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