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Elza [17]
3 years ago
15

What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?

Physics
1 answer:
polet [3.4K]3 years ago
4 0

Answer:

0.044 V

Explanation:

E = Electric field = 5.5\times 10^6\ V/m

d = Thickness of membrane = 8 nm

When the electric field strength is multiplied by the membrane thickness we get the voltage

Voltage across a gap is given by

V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V

The voltage across the membrane is 0.044 V

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A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t
MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity u_x. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

y=u_yt+\frac{1}{2} gt^2

Substitute 0 m/s for u_y, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s

The horizontal distance traveled by the bullet is given by,

x=u_xt

Substitute 500 m/s for u_x and 0.5530s for t.

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The bullet travels a distance of 276.5 m.


5 0
3 years ago
A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500
Neporo4naja [7]

Answer:

A. 4.47 m/s

Explanation:

As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

1500 g = 1.5 kg

E_e = E_k

kx^2/2 = mv^2/2

120*0.5^2/2 = 1.5*v^2/2

15 = 0.75v^2

v^2 = 15 / 0.75 = 20

v = \sqrt{20} = 4.47 m/s

5 0
3 years ago
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