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3 years ago

What is the strength of the electric field 0.020 m from a 12 uc charge?

Physics

1 answer:

belka [17]3 years ago

5 0

Answer:

Option B is correct.

Explanation:

Given the charge, Q = 12uC

Now convert the uC into C = 5*10^-6 C

Given, d = 0.020 m

Now use the below formula:

E = K*Q/d^2

Now insert all the values in the above formula and then solve.

E = (9 ×10^9) × (12×10^-6) / (0.020)^2

E = 2.7 ×10^8 N/C

Therefore option B is correct.

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Name:

Brums [2.3K]

Answer:

1.a) 1 kJ

1.b) 4 kJ

ratio 1:4

1.c) 4 times as before

2.a) 3.33 m/s2

Explanation:

1.a) bicycle's velocity =Displacement/time

=100/20 m/s

=5 m/s

bicycler's KE =1/2 *mass*(velocity)^2

=1/2*80*5^2

=1000 J = 1 kJ

1.b) bicycle's new velocity =200/20 m/s

=10 m/s

bicycler's new KE =1/2*80*10^2

=4000 J = 4 kJ

Ratio= KE 1 :KE new

= 1 :4

1.c) when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it

ex : speed is triple the prior value , then the KE is as 3^2 times as before. that is 9 times

2.a) car acceleration = (20-0)/6 m/s2

= 3.33 m/s2

4 0

3 years ago

What happens to the magnitude of the force of gravitation between two objects if: 1. distance between the objects is tripled? 2.

Ludmilka [50]

<span>2. mass of both objects doubled?
Hope it helped!</span>

8 0

3 years ago

Read 2 more answers

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d

Archy [21]

Answer:

The voltage is $V = 0.993V_b$

Explanation:

From the question we are told that

The time that has passed is $t = \frac{\tau}{2}$

Here $\tau$ is know as the time constant

The voltage of the power source is $V_b$

Generally the voltage equation for charging a capacitor is mathematically represented as

$V = V_b [1 - e^{- \frac{t}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\tau}{2\tau} }]$

=> $V = V_b [1 - e^{- \frac{1}{2} }]$

=> $V = 0.993V_b$

5 0

2 years ago

A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it

Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = $5.93\times 9.81$

T cos θ = 58.17

$T sin \theta =\dfrac{mv^2}{r}$

$T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}$

$T sin^2 \theta =56.93$

$sin^2 \theta = 1 - cos^2 \theta$

$T (1 - cos^2 \theta) =56.93$

$T (1 - (\dfrac{58.17}{T})^2) =56.93$

T² - 56.93T - 3383.75 = 0

T = 93.22 N

$cos \theta = \dfrac{58.17}{93.22}$

θ = 51.39°

6 0

3 years ago

(answer it or get reported) Please answer it w the steps

Travka [436]

Answer;

The mass value for the above kinetic energy equation is 400.0000 kg. This is equal to:

■ 400,000.0000 g.

■ 14,109.6000 ounces.

■ 881.8480 pounds.

3 0

3 years ago