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m_a_m_a [10]
2 years ago
15

When a pitcher throws a baseball, it reaches a top speed of 39 m/s. if the

Physics
1 answer:
Aliun [14]2 years ago
4 0

From the calculation, the acceleration of the body is 26m/s^2.

<h3>What is motion under gravity?</h3>

When an object is thrown up or down, the motion of the body is influenced by the gravitational pull on the body.

Now;

Given that;

v = 39 m/s

t = 1.5 s

u = 0 m/s

a = ?

v = u + at

v = at

a = v/t

a = 39 m/s/1.5 s

a = 26m/s^2

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

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What effect does a parachute have on the frictional force acting on the parachutist?
Schach [20]

Answer: Air resistance and/or drag.

Explanation: The parachute adds drag to the parachutist, thus making him fall slower and safely to the ground.

4 0
3 years ago
the sole of a tennis shoe has a surface area of 0.0290 m^2. if it is worn by a 65.0 kg person, what pressure does the shoe exert
AURORKA [14]

Answer: 21965.517 Pa

Explanation:

Pressure P is the force F exerted by a gas, a liquid or a solid on a surface (or area) A, its unit is Pascal Pa which is equal to N/m^{2} and its formula is:  

P=\frac{F}{A} (1)

In this case we have the surface of a sole of a tennis shoe:

A=0.0290 m^{2} (2)

And the mass m of the person who wears it:

m=65 kg

On the other hand, we know the weight is the force  F the Earth exerts on people and objects due gravity g :

F=m.g=(65 kg)(9.8m/^{2})

F=637N (3)

Substituting (2) and (3) in (1):

P=\frac{637N}{0.0290 m^{2}} (4)

Finally:

P=21965.517 Pa This is the pressure the shoe exert on the ground

5 0
3 years ago
What is the momentum of a photon having the same total energy as an electron with a kinetic energy of 100 keV?
statuscvo [17]

Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000  x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J

Kinetic energy is given as;

K.E = ¹/₂mv²

where;

v is speed of the electron

K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} =  (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} =  (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s

6 0
3 years ago
What is the proper battery cable connection when jumping two automotive batteries? (a) negative to negative / positive to positi
Vladimir [108]
<span>The proper </span><span>battery cable connection when jumping two automotive batteries is :  </span><span>(a) negative to negative / positive to positive. 

</span><span>Connect the red (positive) cable from the car with the bad battery to the red (positive) on the good battery. </span>

<span>Then connect the black (negative) from the good battery to a grounding point on the other car which should be tightened and metal should be clean.
</span>
<span>Once the car with bad battery has started, the removal of the cable should be in the opposite order. The Red (positive) which was the the First Cable to go on should be the last cable to be taken off.</span>


3 0
3 years ago
Let x define the position of an object such that x =
timama [110]
See the attached picture for answers

7 0
2 years ago
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