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disa [49]
3 years ago
12

A fast pitch softball player does a "windmill" pitch moving her hand through a vertical circular arc to pitch a ball at 70mph. T

he 0.19 kg ball is 55 cm from the pivot point at her shoulder.
Just before the ball leaves her hand, what is its centripetal acceleration?
Physics
1 answer:
uysha [10]3 years ago
8 0

Answer:

Explanation:

The hand throwing ball moves on a circular path having radius r = .55 m

mass of ball m = .19 kg

velocity v = 70 mph

= 70 x 1600 / 60 x 60

= 31.11 m /s

centripetal force

= mv² / r

= .19 x 31.11² / .55

= 334.34 N

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A sonar pulse sent out by a boat arrives back after 4 seconds. If the speed of sound in water is 1600m/s, how deep is the water?
Varvara68 [4.7K]

Answer:

the boat would be deeped by 3200 m

Explanation:

Given that

The boat arrives back after 4 seconds

And, the speed of the sound in water is 1,600 m/s

We need to find out how much deep is the water

So,

As we know that

Distance = ( speed × time) ÷ 2

Here we divided by 2 because the boat arrives back

= (1600 × 4) ÷ 2

= 3200 m

Therefore the boat would be deeped by 3200 m

7 0
3 years ago
Difference between reproducibility and replicability
victus00 [196]
Suppose you are doing an experiment where you determine the value of one parameter, say density of a liquid. You have two methods in doing this. By finding the mass and volume, and by using a densitometer. Reproducibility is when you get the same value of density for both methods. Replicability is when you have similar results in one method. So, replicability is a measure of precision, while reproducibility is a measure of accuracy. 
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3 years ago
Describe the difference between contact forces and field forces.
Salsk061 [2.6K]
Contact forces has to be touching for it to be an actual force. A field force does not have to be touching but it does have to be acting on particles at different positions in a space.
8 0
3 years ago
roblem 10: In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following p
miskamm [114]

Answer:

W= -2.5 (p₁*0.0012) joules

Explanation:

Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then

In adiabatic compression, work done by mixture during compression is

W= \int\limits^f_i {p} \, dV  where f= final volume and i =initial volume, p=pressure

p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ

W= \int\limits^f_i {K/V^} \, dV

W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)

W=1/1-γ (p₁Vf-p₀Vi)

W= 1/1-1.40 (p₁*6/5 -p₀*0)  

W= -2.5 (p₁*6/5*0.001)   changing liters to m³

W= -2.5 (p₁*0.0012) joules

3 0
2 years ago
Compare the gravitational acceleration on the following objects compared to the Sun using:
arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765  x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0
2 years ago
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