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katrin [286]
3 years ago
8

A wagon is accelerating to the right. A book is pressed against the back vertical side of the wagon. The coefficient of static f

riction between the book and the vertical back of the wagon is __________ μs.
Physics
1 answer:
sashaice [31]3 years ago
8 0

Answer:

The coefficient of static friction between the book and the vertical back of the wagon is \mu_{s} = mg/N

Explanation:

Since the book is pressed against the vertical side of the wagon, the weight of the book will be acting downwards. i.e. W_{b} = mg

Let the frictional force between the book and the vertical back of the wagon be f_{r}

For the book not to drop down from the back vertical side of the wagon,

f_{r} = mg........................(1)

A static frictional force is given as f_{r} = \mu_{s}  N............(2)

where N= Normal Reaction

Equating (1) and (2)

mg = \mu_{s}  N\\ \mu_{s}  = mg/N

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Corrosion is the irreversible damage or destruction of living tissue or material due to a chemical or electrochemical reaction.
3 0
4 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

and so, substituting:

R=6370 km\\h_A = 5970 km\\h_B = 21200 km

We find

\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}

For satellite A, we have

E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

So, satellite B has the greater total energy (since the energy is negative).

(d) -2.57\cdot 10^8 J

The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

4 0
4 years ago
(physical science) could someone please help me out with this lab? if i’m being honest i did the lab but i lost all of my work :
djverab [1.8K]

Explanation:

hdhhrhhrhehhshsujwuuwuwuwwh

6 0
3 years ago
A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

5 0
3 years ago
Why the distance between two loudspeakers have to be large??​
tangare [24]

Answer:because each speaker has a large angle of of coverage (horizontal and vertical)

They place them apart to prevent their signals (sounds produced) from getting into each other's way as this may cause interference (which may be destructive.

Explanation:

7 0
4 years ago
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