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3 years ago

a sydiver jumps out of a plane. She fall dowward at a very fast speed. Which force is acting upon it?

Physics

1 answer:

UNO [17]3 years ago

6 0

Answer:

There will be two forces acting on her: Gravitational force and Air resisitence

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A 46 g domino slides down a 30 degrees incline at a constant speed. What is the coefficient of friction?

blondinia [14]

Answer:

Explanation:

6 0

3 years ago

Suppose that 2 J of work are needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. How far (in cm) b

sergeinik [125]

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched

<u>Explanation:</u>

Work, W = 2 J

Initial distance, $x_{1}$ = 30 cm

Final distance, = 42 cm

Force, F = 30 N

Stretched length, x = ?

We know,

W = 1/2 kΔx²

Δx = 42-30 cm = 12 cm = 0.12 m

2 J = 1/2 k X (0.12)²

k = 277.77 N/m

According to Hooke's law,

F = kx

30 N = 277.77 X x

x = 0.108 m

x = 10.8 cm

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.

7 0

3 years ago

A woman weighs 129 lb. If she is standing on a spring scale in an elevator that is traveling downward, but slowing up, the scale

Lena [83]

Answer:

C) less than 129 lb.

Explanation:

Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards with magnitude a .

If R be the reaction force

For the elevator is going downwards with acceleration a

mg - R = ma

R = mg - ma

R measures its apparent weight . Spring scale will measure his apparent weight.

So its apparent weight is less than 129 lb .

5 0

3 years ago

The axle of an automobile is acted upon by the forces and couple shown. knowing that the diameter of the solid axle is 32 mm, de

saul85 [17]

First we need to convert the mm to inches to make our computation easier.

1mm = 0.0393701

32mm * 0.0393701 = 1.25 in

Solution:

C = 1/2d = ½ (1.25) = 0.625 in^4

Tension: tension = Te/J = 2T/ piC^3

= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi

Bending:

I = pi/4 * c^4 = 119.842 x 10^-3 in^4

M = (5)(600) = 3600 lb in

G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2 psi = -18.775ksi

Gx = -18.775 ksi

Gy = 0

Txy = 6.519 ksi

G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

1. G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi

G2 = Gave - R = -9/387 - 11.429 = -20.8

Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 = -1.3889

ϴp = -27.1 degrees and 62.9 degrees

2. Tmax = R = 11.43 ksi

R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

3 0

3 years ago

You are on the roof of the physics building, 46.0 above the ground . Your physics professor, who is 1.80 tall, is walking alongs

nekit [7.7K]

There is one mistake in the question as unit of height of building is not given.So I assume it as meter.The complete question is here

You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor’s head, where should the professor be when you release the egg? Assume that the egg is in free fall.

Answer:

d=3.67 m

Explanation:

Height of building=46.0 m

First we need to find time taken by egg to reach 1.80 m above the surface

So to find time use below equation

$S=vt+\frac{1}{2} gt^{2}\\ (46.0-1.80)m=(om/s)t+\frac{1}{2}(9.8m/s^{2} )t^{2}\\t=\sqrt{\frac{(46.0-1.80)m}{4.9} }\\ t=3.06s$

As velocity 1.20m/s is given and we have find time.So we can easily find the distance

$distance=velocity*time\\d=v*t\\d=(1.20m/s)*(3.06s)\\d=3.67m$

3 0

3 years ago