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Tom [10]
3 years ago
11

How does the force block A exerts on block B compare to the force block B exerts on block A?

Physics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

too much force

Explanation:

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A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. T
Mazyrski [523]

Given a capacitor that is charge by a battery

So when the battery was disconnect, the charge remains on the capacitor

One plate will have positive charge

And the other will have a negative charge but of same magnitude

Now, the plates are pulled apart, so the distance between them was extended

a. The capacitance of the capacitor?

The capacitance of a capacitor is given as

C = εo•A/d

Where C is the capacitance

εo is permissivity, a constant

A is are of plate

d is the distance apart

From the formula we notice that

Capacitance is inversely proportional to distance, so this implies that as the distance is increase, the capacitance is reduce.

Capacitance Decreases

b. Potential difference?

The potential difference is given as

q = CV

V = q/C

Or V=Ed

Where E is electric field

So the potential difference is inversely proportional to the capacitance, so we know that the capacitance is reducing, then voltage will increase.

Or, using the second relations

V=Ed, voltage is directly proportional to distance, So, if the distance apart is increasing then, the voltage is increasing

Voltage is increasing

c. Electric field between plate.

The electric field is given as

E = kQ/r²

So the electric field is inversely proportional to the distance square, so as the distance is an increase, the electric field is decreased.

Electric field decreases

d. Electric potential energy?

U = ½CV², then, q = CV

Then, U = ½qV

The electric potential energy is directly proportional to the voltage and since the voltage is increasing due to increase in distance, then the electric potential energy increases.

3 0
2 years ago
A bus slams on its breaks and goes from 30 km/hr to 15km/hr in 4 seconds. What is its acceleration?
PolarNik [594]

Acceleration = (change in speed) / (time for the change).

Change in speed = (15 - 30) = -15 km/hr
Time for the change = 4 sec

Acceleration = (-15 km/hr) / (4 sec) = -3.75 km/hr per second .

Is that a lot ?  Not much ?

Let's convert it to a unit that we can think about:

     (-15 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) =

           (-15 x 1,000) / (3,600)  =  -(4 and 1/6) m/sec .

So the acceleration of the bus is    -(4 and 1/6) m/sec² .

The negative sign means that it slowed down.

(4 and 1/6) m/sec²  is about  42%  of the acceleration of gravity ...
the acceleration the bus would have if it drove off of a cliff.

When the car or the bus you're riding in slows down at that rate,
you feel  42%  of your weight pulling you forward against your
seat belt.  That's quite a drastic acceleration !

5 0
3 years ago
I need help!!!! I don’t understand physical science at all.
allochka39001 [22]
B is the correct answer
3 0
2 years ago
What role did gravity play in the formation of the planets?
joja [24]
Your answer would be D.
If an object has mass, it has gravity, and the more mass it has, the stronger its gravity. During the formation of planets, essentially, various matter and elements pulled and fused together (because of the gravity), forming planetesimals.




3 0
3 years ago
Read 2 more answers
A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes.
Semenov [28]

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

\epsilon=NA\dfrac{dB}{dt}

For N = 1

\epsilon=A\dfrac{dB}{dt}

We need to calculate the current

Using formula of current

i=\dfrac{\epsilon}{R}

Put the value of emf

i=\dfrac{A\dfrac{dB}{dt}}{R}

Now, if the number of turn is 22 , then induced emf would be

\epsilon'=NA\dfrac{dB}{dt}

Then the current would be

i'=\dfrac{\epsilon'}{NR}

i'=\dfrac{NA\dfrac{dB}{dt}}{NR}

i'=\dfrac{A\dfrac{dB}{dt}}{R}

i'=i

Hence, The current would be same in both situation.

4 0
3 years ago
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