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BARSIC [14]
3 years ago
12

the electric current in a wire is 1.5 A. How many electrons flow past a given point in a time of 2 s?​

Physics
1 answer:
34kurt3 years ago
6 0

Answer:

The quantity of electrons that flows past a given point is 3.0 C.

Explanation:

An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).

i.e            I = \frac{Q}{t}

It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.

Given that;     I = 1.5A and t = 2s, find Q.

                          Q = It

                             = 1.5 × 2

                             = 3.0 C

The quantity of electrons that flows past a given point is 3.0 C.

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A box mass of 24kg is being pulled horizontally on a rough surface by an applied force of 585N. The coefficient of kinetic frict
Elan Coil [88]

the normal force is the force applied opposite to the weight of the was box. So the normal force is equal to the weight of the box = 24 kg *(9.81 m/s2) = 235.44 N

the acceleration of the box be solve using newtons 2nd law of motion:

F = ma

a = F/ m = 585 N/ 24 kg = 24.38 m/s2

8 0
3 years ago
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6)the speed of light is approximately​ 186,000 mi/sec. It takes light from a particular star approximately 9 yrs to reach Earth.
NeTakaya

Answer:

5.2791264*10¹³

Explanation:

Convert the 9 years to seconds and then multiple it by 186000

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3 years ago
In which region of the sun does energy move as waves that transfer from atom to atom
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Radiative zone

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ASK YOUR TEACHER A baseball with a mass of 146 g is thrown horizontally with a speed of 40.6 m/s (91 mi/h) at a bat. The ball is
RideAnS [48]

Answer:

Explanation:

mass of the ball = 146 g = 146 / 1000 = 0.146 kg

initial speed of the ball = 40.6 m/s

final speed of the ball = - 45.1 m/s

time of impact = 1.05 ms = 1.05 / 1000 = 0.00105 s

impulse, Ft = change in momentum = mv - mu = m (v-u)

F = m (v - u) / t = 0.146 kg ( -45.1 -40.6) / 0.00105 s = -11916.4 N

4 0
3 years ago
What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force i
CaHeK987 [17]

Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

The net friction force's magnitude will be:

⇒ F_{net}=ma

           =5\times 1.7

           =8.5 \ N

(b)

For m₃,

⇒ ma=\mu m_3 g

Or,

⇒    \mu=\frac{a}{g}

          =\frac{1.7}{9.8}

          =0.173

5 0
3 years ago
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