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Natasha2012 [34]
3 years ago
14

A current exists whenever electric charges move. If ΔQ is the net charge that passes through a surface during a time period Δt,

then the average current during this time interval is defined as average current = ΔQ Δt = Q2 − Q1 t2 − t1 . If we take the limit of this average current over smaller and smaller time intervals, we get what is called the current I at a given time t1: I = lim Δt→0 ΔQ Δt = dQ dt . Thus the current is the rate at which charge flows through a surface. The current in a wire is defined as the derivative of the charge: I(t) = Q'(t). What does b I(t) a dt represent?
Physics
1 answer:
jeka57 [31]3 years ago
5 0

Answer:

It represents the change in charge Q from time t = a to t = b

Explanation:

As given in the question the current is defined as the derivative of charge.

                                  I(t) = dQ(t)/dt ..... (i)

But if we take the inegral of the equation (i) for the time interval  from t=a to

t =b we get

                                   Q =∫_a^b▒〖I(t)  〗 dt

which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.

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What is the horse power of an electric motor which can do by 1250 joule of work in 5 seconds​
Ksju [112]
1250 J in 5 sec= 250 Joule(s) per second (1250/5 0

250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)

250 Watts output = 250/0.65 efficiency = 384 Watts input

1 Horsepower = 732 Watts

Motors 1 Horsepower and under are made in certain step sizes like

3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.

3/4 Horsepower is 549 Watts

1/2 Horsepower is 366 Watts

so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
5 0
2 years ago
A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field
Sophie [7]

Explanation:

direction of electric field is same as that of force experienced by the test charge

8 0
2 years ago
Try this out. Use a calculator!
Maslowich
Just subsitute and easy
v=55m/s
m=100kg
KE=(0.5)(100kg)(55m/s)^2
KE=(50kg)(3025 m^2/s^2)
KE=151250 J
2nd option
6 0
3 years ago
Which event is an example of vaporization?
oksano4ka [1.4K]
B would be an example of vaporization (liquid to gas).

———————

A is an example of deposition (gas to solid); C is an example of condensation (gas to liquid); and D is an example of condensation, deposition, or freezing—depending on the type of cloud.
6 0
3 years ago
A cannon ball is fired directly upward with a velocity of 160 m/s. How long does it take to fall back to the ground? s How fast
Andrej [43]
To answer this problem, we will use the equations of motions.

Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
6 0
3 years ago
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