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Norma-Jean [14]
1 year ago
13

How much charge must pass by a point in a wire in 10 s for the current inb the wire to be 0.50 a?

Physics
1 answer:
Ostrovityanka [42]1 year ago
3 0

Charge must pass by a point in a wire is 5 Coulomb

Given:

time = 10 s = t

current = 0.50 A = I

To Find:

charge = q

Solution: Current is defined as amount of charge passing through a region per unit time. It is given by the formula

I = q/t

q = I x t = 0.50 A x 10 s = 5 Coulomb

Hence, charge must pass by a point in a wire is 5 Coulomb

Learn more about Charge here:

brainly.com/question/25922783

#SPJ4

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Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 320 N/C. (a) What is the ma
nignag [31]

(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².

(b) The speed of the electron after the given time is  4.78 x 10⁵ m/s.

<h3>Acceleration of the electron</h3>

The acceleration of the electron is calculated as follows;

F = qE

ma = qE

a = qE/m

a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)

a = 5.62 x 10¹³ m/s²

<h3>Speed of the electron</h3>

v = at

v = 5.62 x 10¹³ m/s² x  8.50 x 10⁻⁹ s

v = 4.78 x 10⁵ m/s

Learn more about speed here: brainly.com/question/4931057

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7 0
2 years ago
Studen
Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

  D =   72 \  m

b

Magnitude

             d =  - 20 \ m

 Direction  

West

Explanation:

From the question we are told that

The first distance covered westward is d_w_1  =  18 \  m /tex]     The  first distance covered eastward is [tex]d_e1 =  12 \  m /tex]      The second distance covered westward is [tex]d_w_2  =  28 \  m /tex]      The  second distance covered eastward is [tex]d_e2 =  14 \  m /tex]   Generally the distance covered is mathematically represented as       [tex]D =  d_w1 + d_w2 + d_e1 + d_e2

=> D =   18 + 28 + 12 +  14

=> D =   72 \  m

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

d =  -d_w1 -d_w2 + d_e1 + d_e2

=>  d =  -18-28 + 12 + 14

=> d =  - 20 \ m

The direction is west

7 0
3 years ago
2. State Newton's third law of motion.<br>​
Grace [21]

Answer:

Action and reaction are equal but act in opposite directions

4 0
3 years ago
Read 2 more answers
Your mass in kilograms if you weight 170 pounds
Aleonysh [2.5K]
1 pound ≈ 0.4536 kg

170 pounds ≈ 170 * 0.4536 kg

                     ≈ 77.112 kg

                      
5 0
3 years ago
An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,
Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0
3 years ago
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