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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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Answer:

Stephen hawking if his family were scientists

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3 years ago
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How much force must be applied for a 150.W motor to keep a package moving at 3.00m/s?
leva [86]

The force must be applied for motor to keep the package moving is 50 N.

<h3>What is force?</h3>

Force is the action of push or pull the object in order to make it move or stop.

Power is related to force and velocity as

P = F x v

150 W = F x 3 m/s

F  =50 N

Thus, the force must be applied for motor to keep the package moving is 50 N.

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7 0
1 year ago
Estimate frequency of vibration of your arm. Let the length of the arm be 0.57 m. Consider the arm as a simple pendulum and assu
skad [1K]

Answer:

0.80865 Hz

1.23662 seconds

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

l = Length of arm = 0.57 m

Length of simple pendulum is given by

L=\dfrac{2}{3}l\\\Rightarrow L=\dfrac{2}{3}\times 0.57\\\Rightarrow L=0.38\ m

The frequency is given by

f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.81}{0.38}}\\\Rightarrow f=0.80865\ Hz

The frequency is 0.80865 Hz

The time period is given by

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{0.80865}\\\Rightarrow T=1.23662\ s

The time period is 1.23662 seconds

3 0
2 years ago
6. During an impact time casting 5 x 10-45 a gulf club exerts an average impar
Novay_Z [31]

Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

I = F × t

I = 5000 N × 5 × 10-⁴⁵ s

I = 25 × 10-⁴² Ns

I = 2.5 × 10-⁴¹ Ns

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6 0
3 years ago
) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar
ZanzabumX [31]
Momentum would be the same before and after the collision
 Before the collision:
 Momentum of the single cart: 1 * 0.50 = 0.50
 After the collision
 velocity = 0.25m / s
 1 * 0.25 + 1 * 0.25 =
 0.25 * (1 + 1) =
 0.25 * 2 =
 0.50
 Now new momentum will be 0.5
 answer
 the same before and after the collision
4 0
3 years ago
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