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noname [10]
3 years ago
13

How long does it take electrons to get from a car battery to the starting motor? Assume the current is 260 A and the electrons t

ravel through a copper wire with cross-sectional area 0.200 cm2 and length 1.00 m. The number of charge carriers per unit volume is 8.49 × 1028 m-3.
Physics
1 answer:
Digiron [165]3 years ago
4 0

Answer:

17.41 minutes

Explanation:

i = 260 A

A = 0.2 cm^2 = 0.2 x 10^-4 m^2

l = 1 m

n = 8.49 x 10^28 per cubic meter

Let the drift velocity is vd.

The relation between the current and the drift velocity os given by

i = n e A vd

260 = 8.49 x 10^28 x 1.6 x 10^-19 x 0.2 x 10^-4 x vd

vd = 9.57 x 10^-4 m/s

Let the time taken be t

time = distance / speed

t = l / vd = 1 / (9.57 x 10^-4) = 1044.93 seconds = 17.41 minutes

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Explanation:

The protons determines the atomic number which helps in identifying an atom of an element

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A box rests on the back of a truck. The coefficient of friction between the box and the surface is 0.32. (3 marks) a. When the t
cricket20 [7]

Answer:

Part a)

here friction force will accelerate the box in forward direction

Part b)

a = 3.14 m/s/s

Explanation:

Part a)

When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction

Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck

So here friction force will accelerate the box in forward direction

Part b)

The maximum value of friction force on the box is known as limiting friction

it is given by the formula

F = \mu mg

so we have

F = ma = \mu mg

now the acceleration is given as

a = \mu g

a = (0.32)(9.8) = 3.14 m/s^2

5 0
3 years ago
What does increasing the impulse on an object due to its momentum?
allochka39001 [22]

increaskng the impulse also <u>increases </u><u>momentum</u>

7 0
2 years ago
3. Two cyclists that are 500 m apart start biking toward each other. They bike at speeds of 6 and
Veronika [31]

Answer:

A) 50 seconds

B) 200 m

Explanation:

They are 500 metres apart.

And one of the bike loves at 6 m/s while the other loves at 4 m/s.

A) Let distance of the 6 m/s bike before they meet be x.

Thus, time = x/6

Since time = distance/speed

For the second bike at 4 m/s, his distance covered before they meet will be 500 - x

Thus, time = (500 - x)/4

Now they will meet each other at the same time. Thus;

x/6 = (500 - x)/4

Cross multiply to get;

4x = 3000 - 6x

6x + 4x = 3000

10x = 3000

x = 3000/10

x = 300 m

Thus, time will be;

t = 300/6

t = 50 seconds

B) Distance covered by the slower bike is (500 - x)

Since from a above, x = 300

Thus; distance = 500 - 300 = 200 m

4 0
3 years ago
If the range of the projectile is 4.3 m, the time-of-flight is T = 0.829 s, and air resistance is negligible, determine the foll
ankoles [38]

Answer

given,

range of the projectile = 4.3 m

time of flight = T = 0.829 s

v =\dfrac{d}{T}

v =\dfrac{4.3}{0.829}

     v = 5.19 m/s

vertical component of velocity of projectile

v_y = gt'

v_y = 9.8 \times {\dfrac{T}{2}}

v_y = 9.8 \times {\dfrac{0.829}{2}}

v_y =4.06\ m/s

a) Launch angle

 \theta = tan^{-1}(\dfrac{v_y}{v})

 \theta = tan^{-1}(\dfrac{4.06}{5.19})

    θ = 38°

b) initial speed of projectile

  v = \sqrt{v_x^2 + v_y^2}

  v = \sqrt{5.19^2 + 4.06^2}

         v = 6.59 m/s

c) maximum height reached by the projectile

     y_{max}=v_{avg}t'

     y_{max}=\dfrac{1}{2}v_y\dfrac{T}{2}

     y_{max}=\dfrac{1}{2}\times(g\dfrac{T}{2})\times\dfrac{T}{2}

     y_{max}=\dfrac{gT^2}{8}          

7 0
4 years ago
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