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nikklg [1K]
3 years ago
12

john is using a pulley to lift the sail on his sailboat. the sail weighs 150N and he must lift it at 4.0m. how much work must be

done on the sail?
Physics
1 answer:
gogolik [260]3 years ago
5 0

The work done on the sail is 600 J

Explanation:

The work done to lift the sail is equal to the gain in gravitational potential energy of the sail, therefore is:

W=mg\Delta h

where

m is the mass of the sail

g is the acceleration of gravity

(mg) is the weight of the sail

\Delta h is the change in height of the sail

In this problem we have

mg = 150 N (weight)

\Delta h = 4.0 m

Substituting, we find the work done:

W=(150)(4.0)=600 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Answer:

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Explanation:

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What is the potential difference (voltage) if the resistance of 20 ohms produces a current of 10 amperes?
JulijaS [17]

Answer:

200 volts

Explanation:

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2 years ago
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2 years ago
022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati
svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

v_f = 24.5 m/s

Explanation:

Part 1)

initial speed of the ball upwards

v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

H = \frac{24.5^2}{2(9.80)}

H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

0 - 24.5 = (-9.8)t

t = 2.5 s

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

v_f = 24.5 m/s

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

y = \frac{1}{2}at^2

4.88 = \frac{1}{2}a(1^2)

a = 9.76 m/s^2

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

a = 90 m/s^2

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

v_f = v_i + at

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v_f = 29.7 m/s

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}(9.81)(2.3^2)

y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

d = 2.92 \times 2.3

d = 6.72 m

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Answer:

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So speed =\frac{distance}{time}=\frac{60}{6}=10km/hr

So average speed of car in the first trip is 10 km/hr

7 0
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