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4 years ago

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A 1 kg ball moving horizontally to the right at 3 m/s strikes a wall and rebound, moving horizontally to the left at the same ve

locity of 3 m/s. What is the magnitude of the change in linear momentum of the ball?

1 answer:

Airida [17]4 years ago

3 0

Answer:

The magnitude of the change in linear momentum of the ball is 6 kg-m/s.

Explanation:

Given that,

Mass of the ball, m = 1 kg

Initial sped of the ball, u = 3 m/s

It strikes a wall and rebound, moving horizontally to the left at the same velocity of 3 m/s, v = -3 m/s

The magnitude of the change in linear momentum of the ball is given by :

$p=m(v-u)\\\\p=1\ kg\times (-3-3)\\\\p=-6\ kg-m/s$

So, the magnitude of the change in linear momentum of the ball is 6 kg-m/s.

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If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *

Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

$F_s = \mu_s N$

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

$F_k = \mu_k N$

now we know that

$\mu_k < \mu_s$

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

so kinetic friction may be greater than 400 N or smaller than 400 N

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3 years ago

Which of the following is not true?

ahrayia [7]

The answer is either C or D..

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3 years ago

Physical science semester B exam 66 problems , please don’t delete I’m on my last leg I need this please god help me

krek1111 [17]

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4 years ago

A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around

Alex_Xolod [135]

Answer:

$\alpha =\frac{m*g*R}{I-m*R^2}$

$a = \frac{m*g*R^2}{I-m*R^2}$

$T=\frac{I*m*g}{I-m*R^2}$

Explanation:

By analyzing the torque on the wheel we get:

$T*R=I*\alpha$ Solving for T: $T=I/R*\alpha$

On the object:

$T-m*g = -m*a$ Replacing our previous value for T:

$I/R*\alpha-m*g = -m*a$

The relation between angular and linear acceleration is:

$a=\alpha*R$

So,

$I/R*\alpha-m*g = -m*\alpha*R$

Solving for α:

$\alpha =\frac{R*m*g}{I+m*R^2}$

The linear acceleration will be:

$a =\frac{R^2*m*g}{I+m*R^2}$

And finally, the tension will be:

$T =\frac{I*m*g}{I+m*R^2}$

These are the values of all the variables: α, a, T

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3 years ago

A satellite of Mars, called Phobos, has an orbital radius of 9.4 ✕ 106 m and a period of 2.8 ✕ 104 s. Assuming the orbit is circ

FromTheMoon [43]

Answer:

$6.27*10^{23}kg$

Explanation:

assume

M= mass of Mars

m=mass of phobos

r=orbital radius

T=period

we can apply F=ma to this orbital motion (considering the cricular motion laws)

where,

$F=\frac{GMm}{r^{2} }$ and a=rω^2

where ω= $\frac{2\pi }{T}$ and G is the universal gravitational constant.

G = 6.67 x 10-11 N m2 / kg2

$F=ma\\\frac{GMm}{r^{2} }=mr(\frac{2\pi }{T} )^{2}\\ M=\frac{r^{3}}{G} (\frac{2\pi }{T} )^{2}\\M=\frac{(9.4*10x^{6} )^{3}*(2\pi )^{2} }{(2.8*10^{4}) ^{2} *6.67*10^{-11} } \\M=6.27*10^{23}kg$

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3 years ago