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timurjin [86]
3 years ago
14

A batter hits two baseballs with the same force. One hits the ground near third base. The other is a home run out of the park. W

here was there more work done? Explain your answer.
The ball that went out of the park shows more work because the force applied was greater.

The ball that hit near third base shows more work because it hit the ground much harder.

The ball that went out of the park shows more work because the distance was greater.

The ball that hit near third base shows more work because the force applied was greater.
Physics
1 answer:
Basile [38]3 years ago
3 0
I think the answer is "<span>The ball that went out of the park shows more work because the distance was greater."</span>
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When a generator rotates a coil of wire in a magnetic field, which of the following is produced?
Sloan [31]
The correct answer is an ''electric current''.
4 0
3 years ago
Which columns are mislabeled?
Aleks [24]

Answer:

first order date and most recent order date

Explanation:

it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019

8 0
3 years ago
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A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp
Anna71 [15]

\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}

Explanation:

       Natural length of a spring is 0.5\text{ }m. The spring is streched by 0.02\text{ }m. The resultant energy of the spring is 0.5\text{ }J.

       The potential energy of an ideal spring with spring constant k and elongation x is given by \dfrac{1}{2}kx^{2}.

       So, in the current problem, the natural length of the spring is not required to find the spring constant k.

       \text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}

∴ The spring constant of the spring = 2500\text{ }\frac{kg}{s^{2}}

4 0
4 years ago
A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the
Mariana [72]

Answer:

17.72° or 72.28°

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u = 6.5 m/s

R = 2.5 m

Let the angle of projection is θ.

Use the formula for the horizontal range

R=\frac{u^{2}Sin2\theta }{g}

2.5=\frac{6.5^{2}Sin2\theta }{9.8}

Sin 2θ = 0.58

2θ = 35.5°

θ = 17.72°

As we know that the range is same for the two angles which are complementary to each other.

So, the other angle is 90° - 17.72° = 72.28°

Thus, the two angles of projection are 17.72° or 72.28°.

8 0
4 years ago
A dry cell does 7.5 j of work through chemical energy transfer 5.00C between terminals of the cell . What is the electric potent
lisov135 [29]
Here's a useful factoid that you don't hear about very often:

1 volt means 1 Joule per Coulomb.

When 1 coulomb of charge falls or gets lifted through 1 volt potential difference, it gains or loses 1 Joule of energy.

If you want to lift 5 coulombs to a height of 1 volt, you have to give it 5 joules.

If you actually give those 5 coulombs 7.5 joules instead, they'll rise up to 1.5 volts above the potential where they started. The flowed through a potential DIFFERENCE of 1.5 volts.

(If they started at a point that's connected to the Earth, like a water pipe or a metal flagpole, then their new potential is 1.5 volts, because we define zero as the potential of the ground.)
7 0
3 years ago
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