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MakcuM [25]
3 years ago
7

How do you find the coefficient of friction?

Physics
1 answer:
DaniilM [7]3 years ago
5 0

Answer:

There's a formula for finding the coefficient of friction i.e. μ = f÷N

hope it helps!

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A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 10.5 m/s.
lesantik [10]

Answer:

Height as seen by the professor = 38.2 m

Explanation:

Angle of throw = θ = 69°

Velocity of throw = v

X component of velocity = v₁ = v cos 69 = 0.3584 v m/s

Vertical component of the velocity = v₂ =  v sin 69 = 0.9336 v m/s

v₂ / v₁ = tan 69 = 2.605

v₂ = 2.605 v₁.

Professor sees as if the x component of velocity =0

v (as seen by professor) + v'  = 0

=> v  as seen by professor = -v' = -10.5 m/s

This shows that y component of the ball's velocity is 2.605 times its x component of velocity.

with respect to the professor, there is only y component of velocity.

v₂' =v₂ = 2.605 ( -10.5) = 27.4 m/s.

Height as seen by the professor = (27.4)² / 2(9.8) = 38.2 m

4 0
4 years ago
Develop expression for equation for uniform motion in a straight line
frozen [14]
According to cuneiform tablets in the ancient world, straight lines cannot cross, and no motion in the world is not relative. Btw...I KNOW!!! GOT MILK???
4 0
3 years ago
A car is moving eastward and speeding up. The momentum of the car is ______.
leva [86]
I think the answer is C. Remaining constant
7 0
3 years ago
For a wave, what term is defined as the maximum height of a crest, or depth of a trough, relative to the normal level? for a wav
sammy [17]
The appropriate response is amplitude. Amplitude is the target estimation of the level of progress (positive or negative) in environmental weight (the pressure and rarefaction of air atoms) created by sound waves.

I hope the answer will help you. 
6 0
3 years ago
Two ions with masses of 5.29×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is
riadik2000 [5.3K]

Answer:

0.132 m

Explanation:

m = mass of the ion = 5.29 x 10⁻²⁷ kg

q = magnitude of charge on singly charged ion = 1.6 x 10⁻¹⁹ C

r = radius of circular path followed by singly charged ion

v = speed of the ion = 1.13 x 10⁶ m/s

B = magnitude of the magnetic field = 0.283 T

Radius of the circular path is given as

r = \frac{mv}{qB}

r = \frac{(5.29\times 10^{-27})(1.13\times 10^{6})}{(1.6\times 10^{-19})(0.283)}

r = 0.132 m

6 0
3 years ago
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