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andrey2020 [161]
3 years ago
6

Question below...............

Physics
1 answer:
kondaur [170]3 years ago
7 0

Answer:

friction force

Explanation:

force of friction is opposite to the force applied it resist the motion

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What is the term of movement in a particular direction
Art [367]
That is a vector. It is a combination of direction and velocity. (You can think of Vector from Despicable Me to help you remember the term)
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3 years ago
Two large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 170 V. An electron ent
notsponge [240]

Answer:

Explanation:

Electric field between plates

V / d

= 170 / ( 2 x 10⁻² )

= 8500 N/C

Force on electron in this field

= 8500 x 1.6 x 10⁻¹⁹

= 13600 x 10⁻¹⁹ N

Acceleration

= 13600 x 10⁻¹⁹ / 9.1 x 10⁻³¹

a = 1494.5 x 10¹² m /s²

s = .1 x 10⁻² m

v² = u² + 2as

=  (2.9x 10⁵)²+ 2 x 1494.5 x 10¹² x .1 x 10⁻²

= 8.41 x 10¹⁰ + 299 x 10¹⁰

= (8.41 + 299 ) x 10¹⁰

v = 17.53  x 10⁵ m /s

8 0
3 years ago
If you were on a ship at sea, and a tsunami passed under your ship, what would probably be your reaction? explain.
swat32
<span>Extremely powerful single waves have no effect on ships at sea since the depth of water allows the energy to be distributed over hundreds and thousands of feet. In deep water, the bigger the wave, the faster it moves and the slower the surface changes height. As the wave gets into shallow waters, it slows down and can start to pile up to large heights.</span>
6 0
3 years ago
Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The forc
erastova [34]
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3 0
3 years ago
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
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