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AveGali [126]
3 years ago
8

If the current in a wire of a cd player is 12.0 ma, how long would it take for 6.50 c of charge to pass a point in this wire? an

swer in units of s.
Physics
1 answer:
Galina-37 [17]3 years ago
6 0
The current intensity is defined as the amount of charge Q that passes through a certain point of an electrical wire in a time interval of \Delta t:
I= \frac{Q}{\Delta t}

In our problem, the current intensity is
I=12.0 mA=0.012 A
while the amount of charge that passes a certain point of the wire is 
Q=6.50 C

If we re-arrange the previous equation:
\Delta t= \frac{Q}{I}
we can find the time needed for this amount of charge to pass through a point of the wire:
\Delta t= \frac{6.50 C}{0.012 A} =541.7 s
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A student wearing a frictionless roller skates on a horizontal is pushed by a friend with a constant force of 55N. How far must
umka21 [38]

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

362 = 55d

d = 362/55

d = 6.58m

Hence the student must push at a distance of 6.58m

3 0
3 years ago
At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at
noname [10]
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3 0
3 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
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