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Lyrx [107]
3 years ago
11

Pls help its about baseball and energy (one question) '^'

Physics
1 answer:
avanturin [10]3 years ago
5 0
The answer would be A. As the bat is swung, it gains kinetic energy. But once it hits the ball, it loses, or transfers, it’s kinetic energy to the ball.
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what branch of Science studies matter and energy and the way they work together with force to cause motion
algol13
The branch of science that studies this is physics

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8 0
3 years ago
2. A 75 kg runner accelerates from 0.00 m/s to 10.0 m/s in 1.5 seconds.
raketka [301]
A :-) A.a.) Given - u = 0.00 m/s
v = 10.0 m/s
t = 1.5 sec
m = 75 kg
Solution -
a = v - u by t
a = 10 - 0 by 1.5
a = 10 by 1.5
a = 6.6 m/s^2

A.b.) sorry ! I don’t no how to do this question
3 0
3 years ago
If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re
-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = \frac{1}{2}kx^2   ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = \frac{1}{2}k\times0.021^2

or

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = \frac{1}{2}\times13605.44\times (0.062)^2

or

U = 26.149 J

Also,

Work done = change in energy

or

W = 26.149 - 3.0 = 23.149 J

4 0
3 years ago
Which of the following would have the smallest gravitational attraction between the two masses?
expeople1 [14]

Answer: Distance= 100,000 km

   Mass= 15 million kg        Mass= 5 million kg

3 0
3 years ago
Read 2 more answers
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
3 years ago
Read 2 more answers
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