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prisoha [69]
3 years ago
9

The switch on the electromagnet, initially open, is closed. What is the direction of the induced current in the wire loop (as se

en from the left)?
Physics
1 answer:
Marianna [84]3 years ago
6 0

Answer:

The induced current is clockwise

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approximation to the average velocity in that time interval, what should be the sequence of calculations?Update the (vector) pos
Klio2033 [76]

Answer:

The steps are outlined in the explanation below.

Explanation:

The average velocity is derived midpoint from the initial to the final velocity. Here is the proof:

Find the total displacement:

let the displacement be given by the letter s

Then since the average velocity is defined as:  v_{av}  = \frac{x - x_{0} }{t - t_{0} }

where t = final time

           t₀ = initial time

           v = final speed

           v₀ = initial time

where x denotes the position, then

v_{ave} = \frac{v+v_{0} }{2}

where v = \frac{dx}{dt} and dx = change in distance with respect to time.

6 0
3 years ago
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
What is the momentum of a 1400 kg car traveling at 25 m/s?
sveticcg [70]
<h2>Hello</h2>

The answer is:

35000Kg.\frac{m}{s}=35000N

<h2>Why?</h2>

Momentum is the quantity of movement of an object, and it's calculated using the mass and the velocity of the object. Momentum is expressed by the following formula:

p=m*v

Where:

m=mass\\v=velocity

So, calculating we have:

p=m*v=1400Kg*\frac{25m}{s}=35000Kg.\frac{m}{s}=35000N

Remember,

1N=1Kg.\frac{m}{s}

Have a nice day!

5 0
2 years ago
Read 2 more answers
What happens to light when it strikes the inside surface of a smooth, curved mirror? It bounces in random directions. It bounces
kotykmax [81]
This is a concave mirror you're talking about, so all of the points are going to converge to a single focal point. Therefore the answer would be that it bounces back toward a single spot
5 0
3 years ago
Read 2 more answers
Someone answer pls !!!!
astra-53 [7]

Answer:

My best guess would be B due to the fact of friction in a simple machine

7 0
3 years ago
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