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aalyn [17]
3 years ago
13

Calculate the force between charges of 55 × 108 C and 1 × 107C if they are 5 cm apart.?​

Physics
1 answer:
lions [1.4K]3 years ago
6 0

First:

5 cm = 0,05 m

Now, the formula:

F = k * (qq') / r²

Replacing:

F = 9x10⁹ * (55x10⁸ * 1x10⁷) / 0,05²

F = 9x10⁹ * (5,5x10¹⁶ / 2,5x10⁻³)

F = 9x10⁹ * 2,2x10¹⁹

F = 1,98x10²⁹

The force between the charges is 1,98x10²⁹ Newtons.

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A student who was training for a cross country race jogged for 2. 0 hours and covered a distance of 14. 0 kilometers. What was t
FinnZ [79.3K]

The average speed is 116.66m/s.

Given - The path traced is 14km , time for jogging is 2 hrs=120min

To find the average speed-

  1. Speed refers to the ease of the movement and degree of mobility as a result of force application.
  2. Due to this there is involvement of velocity.
  3. Journey of average speed is the cumulative of distances and time.
  4. Kinetic theory refers to the Boltzmann constant connecting to the standards of distance traversed.

calculations-

Speed= \frac{Distance}{Time}

Speed= 14 000 / 120

= 116.66m/s

To learn more about average speed -

<u>brainly.com/question/27753148</u>

#SPJ4

6 0
1 year ago
A sample contains 36 g of a radioactive isotope. How much radioactive isotope remains in the sample after 3 half-lives?
Gre4nikov [31]

Answer:

<u>Option "C":</u> "4.5 g"

Explanation:

N0 = 36 g, Let half-life is T.

t = 3 T, n is number of half lives = t / T = 3

<u>By using the decay law of radioactivity</u>

N / N0 = (1 / 2)^n

where

"N0" be the "initial amount"

"N" be the "amount left"

"n" be the "number of half-lives"

N / 36 = (1/2)^3

N / 36 = 1 / 8

N = 36 / 8 = 4.5 g

3 0
2 years ago
Read 2 more answers
Glider one and glider two collided. The data table above shows the momentum of each before and after the collision. Perform an a
k0ka [10]

I think there was momentum conserved

Explanation: I took the test

7 0
3 years ago
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward.
fgiga [73]

complete question:

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answer:

F = 1776  N

Explanation:

mass of ball = 60 g = 0.06 kg

velocity of downward direction = 22 m/s = v1

velocity of upward direction = 15 m/s = v2

Δt = 1/800 = 0.00125 s

Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.

p = mv

When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .

I = pf − pi = ∆p

F =  ∆p/∆t  =  I/∆t

let the upward velocity be the positive

Δp =  mv2 - m(-v1)

Δp =  mv2 - m(-v1)

Δp = m (v2 + v1)

Δp = 0.06( 15 + 22)

Δp = 0.06(37)

Δp = 2.22 kg m/s

∆t  = 0.00125

F =  ∆p/∆t

F =  2.22/0.00125

F = 1776  N

4 0
3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
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