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Alona [7]
4 years ago
7

A 0.75μF capacitor is charged to 70 V . It is then connected in series with a 55Ω resistor and a 140 Ω resistor and allowed to d

ischarge completely. Part A How much energy is dissipated by the 55Ω resistor?
Physics
1 answer:
Ipatiy [6.2K]4 years ago
6 0

Answer:

Explanation:

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.

So for 55ohms, using voltage divider rule

V=R1/(R1+R2) ×Vt

V=55/(55+140) ×70

V=19.74Volts is across the 55ohms resistor.

Then, energy loss will be

E=V²/R

E=19.74²/55

E=7.09J

7.09J of heat is dissipated by the 55ohms resistor

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Example:

A rock vs a feather
The rock will fall quicker because it’s more dense and falls straight down and the feather will be slower because it flows slowly down through the air particles
6 0
3 years ago
A skateboarder, starting from rest, rolls down a 10.9-m ramp. When she arrives at the bottom of the ramp her speed is 6.74 m/s.
JulijaS [17]

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4 0
3 years ago
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 30.0 mph and half the
salantis [7]

Answer:

Explanation:

Given

Distance to grandmother's house=100 mi

it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph

Let Julie travel x mi with 30 mph and 100-x with 70 mph

\frac{x}{30}=\frac{100-x}{70}

x=30 mi

Therefore

Julie's Average speed on the way to Grandmother's house=\frac{100}{\frac{50}{30}+\frac{50}{70}}

=42 mph

On return trip

=\frac{100}{2\frac{30}{30}}=50 mph

6 0
4 years ago
A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi
Olenka [21]
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

At the start:

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8 0
3 years ago
WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!
faust18 [17]

Answer:

9)a

10) I think true

11)b

Explanation:

9)a. because it's told that the car is slowing down, the sum of the forces that are towards left, should be more than the ones that are towards right. if the car was gaining speed, "b" would have been correct. and if it was told that the car is moving without a change in the speed, "c" would have been correct.

10) if a moving object has a change of speed or direction, it would have an acceleration. now if a moving object experiences an unbalanced force, it'd either slow down, gain speed or change direction, and in all of the three possibilities it'd have an acceleration.

11) upward and downward forces are equal, and the sum of them would be 0N(because they have opposite directions). so they negate each other.

and the rightward force is 5N more than the leftward force. so the Net Force would be 5N.

-30+30-10+15=5N

if it is unclear or you need more explanation, ask freely.

8 0
3 years ago
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