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3 years ago

How do we know Earth is spinning, not the Sun?

Physics

1 answer:

Tpy6a [65]3 years ago

4 0

Answer:

Because it is said that the earth rotates and revolves around the sun and moon so it is impossible for the earth not to be spinning.

Explanation:

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Find the horizontal component and the vertical component

aleksandr82 [10.1K]

Answer:

v=3.66,h-3.66

Explanation:

vertical = 10sin60 - 10sin 30

horizontal =10cos60 + 10cos 30

v = 10×0.8660-10×0.5

h = 10×0.5 + 10 × 0.8660

v=8.660-5.0 = 3.66

h= 5.0-8.660 = -3.66

8 0

2 years ago

Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s

Inga [223]

The average force applied to the ball= 106.7 N

Explanation:

Force is given by

f= ΔP/t

ΔP= change in momentum= m Vf- m Vi

m= mass =0.2 kg

Vf= final velocity= 12 m/s

Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)

t= time= 60 ms= 0.06 s

now ΔP= 0.2 [ 12-(-20)]

ΔP=0.2 (32)=6.4 kg m/s

now force F= ΔP/t

F= 6.4/0.06

F=106.7 N

8 0

3 years ago

Hi :) how to do this?

fredd [130]

Answer:

The answer is b

Hope this helps you!

4 0

3 years ago

A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on

Juliette [100K]

Answer:

k = $5\times 10^{4}\ N/m$

b = $0.707\times 10^{3}$

t = $7.1\times 10^{- 5}\ s$

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

$v^{2} = u^{2} + 2gh$

where

u = initial velocity = 0

g = 10 $m/s^{2}$

Thus

$v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s$

Force on the mass is given by:

F = mg = $1000\times 10 = 10000 N = 10\ kN$

Also, we know that the spring force is given by:

F = - kx

Thus

$k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m$

Now, to find the damping constant b, we know that:

F = - bv

$b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}$

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

$t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s$

4 0

3 years ago

A force is applied to a block to move it up a 30° incline. The incline is frictionless, the block is initially at rest, F= 65.0

muminat

Answer:

0.96 m, upward

Explanation:

$\theta=30^{\circ}$

$F=65 N$

$M=5 kg$

Initial velocity, u=0

$t=0.550 s$

$Fcos\theta-Mgsin\theta=Ma$

Where $g=9.8 m/s^2$

Substitute the values

$65cos30-5\times 9.8sin30=5a$

$31.8=5a$

$a=\frac{31.8}{5}=6.36 m/s^2$

$S=ut+\frac{1}{2}at^2$

$S=0+\frac{1}{2}(6.36)(0.55)^2=0.96 m$

Hence,the block moves upward because displacement is positive.

7 0

3 years ago