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Levart [38]
2 years ago
8

You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The

bucket is attached to a heavy duty steel chain. The mass of the chain is 19.3 kg.
How much work do you perform during the lifting process?

8427 J is Incorrect.

If it takes 1.75 minutes for you to raise the bucket of water out of the well, then what was your average power?

80.25 W is Incorrect.
Physics
1 answer:
Mars2501 [29]2 years ago
3 0

The total work done is  5980 Joules and the power expended is 57 Watts.

<h3>What is work done?</h3>

The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

Height of the center of mass of chain = 25.9 / 2 = 12.95 m  

Work done by the chain Wc;

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules  

Total work = 3530 + 2450 = 5980 Joules

Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

Learn more about work done:brainly.com/question/13662169

#SPJ1

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A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c
Anna35 [415]

The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

The direction of the acceleration of the ball is downwards

The given parameters

initial velocity of the ball, u = 0

height above the ground, h = 2.2 m

time of motion of the ball, t = 96 ms = 0.096 s

The magnitude of the acceleration of the ball while coming to rest is calculated as;

let the downwards direction of the acceleration be positive

h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2

The direction of the acceleration of the ball is downwards

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4 0
2 years ago
Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.
Vesnalui [34]

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

ma = \text{Mass}\times \text{Acceleration}

ma = kg \cdot \frac{m}{s^2}

At the same time we have that

\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2

\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2

\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}

Therefore there is not have same units and both are not consistent and the correct answer is B.

5 0
3 years ago
Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

8 0
3 years ago
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