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Darya [45]
3 years ago
5

Which celestial body would have the strongest gravitational pull on a satellite orbiting 100 km above its surface? Explain your

answer. pls do this in your own words this is for egenuity and don't use the sample response
Physics
1 answer:
Nostrana [21]3 years ago
6 0
So the exam was about 300 credits
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The amplitude of the electric field is 0.9 V/m 42 km from a radio transmitter. What is the total power emitted by the transmitte
Nadusha1986 [10]

Answer:

The power is found to be P = 14250600 W

Explanation:

First we find the voltage or potential difference at the position:

Voltage = Amplitude of Field x Distance

Voltage = V = (0.9 V/m)(42000 m)

V = 37800 volts

Now, we know the formula of electric power is:

Power = Voltage x Current

P = V I

but, from Ohm's Law:

V = I R

I = V/R

Therefore,

P = V²/R

Resistance = real part of impedence = 377 Ω

Therefore,

P = (37800 V)(377 Ω)

<u>P = 14250600 W</u>

7 0
3 years ago
Will give brainly!! Help pls!!!
Naily [24]
We use friction to create warmth when we run our hands together.
it helps us from slipping on the floor. that's the only reason we can walk without falling. there is way less friction on wet floor, due to which we fall
5 0
3 years ago
The effects due to the interaction of a current-carrying loop with a magnetic field have many applications, some as common as th
antoniya [11.8K]

Answer:

a) τ = 1.039*10⁻⁴N-m

b) The net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise  direction.

Explanation:

A) Given

I = 0.5 A

B = 0.3 T

a = 4 cm = 0.04 m

b = 2 cm = 0.02 m

θ = 30°

The torque τ acting on a current-carrying loop of area A due to the interaction of the current I flowing  through the loop with a magnetic field of magnitude B is given by

τ = I*B*A*Sin∅

where ∅ is the angle between the normal to the loop and the direction of the magnetic field.

The area A of a rectangular loop of wire with height 4.00 cm and horizontal sides 2.00 cm can be obtained as follows

Aloop = a*b   ⇒   Aloop = 0.04 m*0.02 m = 8*10⁻⁴m²

Recalling that θ is the angle between the  sides of length b and B and if we consider the normal to the loop, the angle ∅ between the normal  and the magnetic field is given by

∅ = 90°-θ    ⇒   ∅ = 90°-30° = 60°

Then, the torque will be

τ = (0.5 A)*(0.3 T)*(8*10⁻⁴m²)*Sin60° = 1.039*10⁻⁴N-m

b) We have to get the net torque τ about the vertical axis of the current loop due to the interaction of the current  with the magnetic field.

The angle ∅ between the normal to the loop and the magnetic field when the  horizontal sides of the loop of length b are perpendicular to B is

∅ = 0°

Then

τ = (0.5 A)*(0.3 T)*(8*10⁻⁴m²)*Sin 0° = 0 N-m

We can say that the net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise  direction.

5 0
4 years ago
Why is there no electric field at the center of a charged spherical conductor?
Alja [10]
I reported this question by accident, I am sorry.

Anyways, there is no electric field at the center of a charged spherical conductor because all the charged particles are at the edges and because of the complete cancellation everywhere.
5 0
3 years ago
You want to use the idea of electromagnetic induction to make the bulb in your small flashlight glow; it glows when the potentia
makkiz [27]

a) 57.5 m/s

b) Yes

Explanation:

a)

According to Faraday-Newmann-Lenz's law, the electromotive force induced in the coil due to the change in magnetic flux through it is given by:

\epsilon=-\frac{N \Delta \Phi}{\Delta t}

where

N is the number of turns in the coil

\Delta \Phi is the change in magnetic flux

\Delta t is the time interval

The change in magnetic flux can be written as

\Delta \Phi = A\Delta B

where

A is the area of the coil

\Delta B is the variation of the strength of the magnetic field

Re-writing the equation,

\epsilon=-\frac{NA\Delta B}{\Delta t}

To make the bulb glowing, the induced emf must be:

\epsilon=1.5 V

And we also have:

N = 100

\Delta B=0 T-3.6\cdot 10^{-2} T=-0.036 T

A=2.9\cdot 10^{-4} m^2

So we can find the maximum time required to induce this emf:

\Delta t=-\frac{NA\Delta B}{\epsilon}=-\frac{(100)(2.9\cdot 10^{-4})(-0.036)}{1.5}=6.96\cdot 10^{-4} s

Since the length to cover in this time is

L = 4.0 cm = 0.04 m

The speed should be

v=\frac{L}{t}=\frac{0.04}{6.96\cdot 10^{-4}}=57.5 m/s

b)

Yes: if the coil is moved at a speed of 57.7 m/s, then the potential difference induced in the bulb will be 1.5 V, which is enough to make the bulb glowing.

6 0
3 years ago
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