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natita [175]
3 years ago
6

what is the numerical value in meters per second squared of the acceleration of an object experiencing true free fall

Physics
1 answer:
Rina8888 [55]3 years ago
8 0

Answer:

9.8m/s²

Explanation:

An object experiening a free fall is simply under the influence of gravity.

When an object is influenced by gravity, the object falls with an acceleration known as acceleration due to gravity.

This acceleration known as acceleration due to gravity has a constant value of 9.8m/s².

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A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball
RideAnS [48]

Answer:

C=2.32\times 10^{-4}\ Ns^2/m^2

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, F_r=Cv^2

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

Cv^2=mg

C=\dfrac{mg}{v^2}

C=\dfrac{0.046\times 9.8}{(44)^2}

C=2.32\times 10^{-4}\ Ns^2/m^2

Hence, this is the required solution.

4 0
3 years ago
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.40 m/s
tekilochka [14]

Answer:

a) 378Ns

b) 477.27N

Explanation:

Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula

I = Ft = m(v-u)

m is the mass = 70kg

v is the final velocity = 5.4m/s

u is the initial velocity = 0m/s

Get the impulse

I = m(v-u)

I = 70(5.4-0)

I = 70(5.4)

I = 378Ns

b) Average total force is expressed as

F = ma (Newton's second law)

F = m(v-u)/t

F = 378/0.792

F = 477.27N

Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N

3 0
3 years ago
Lunar missions have revealed that the moon has:
myrzilka [38]
That the moon has soil within its shadowy craters rich and useful material
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In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a
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<span>We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
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Analyze how some insects are able to move around on the surface of a lake or pond
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