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lorasvet [3.4K]
3 years ago
5

A 4.0-μF capacitor that is initially uncharged is connected in series with a 4.0-kΩ resistor and an ideal 17.0-V battery. How mu

ch energy is stored in the capacitor 17 ms after the battery has been connected?
Physics
1 answer:
Komok [63]3 years ago
4 0

Answer:

Indian girls

Explanation:

I don't know what do you expect I don't know this

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What is the angle of reflection of the incident angle is 35°?<br> 35<br> 25°<br> 55<br> 90
Anestetic [448]

Answer:

i think its the third one or second one

4 0
2 years ago
local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength at which they are broadcasting. Waveleng
natta225 [31]

Answer:

λ = 437 m.

Explanation:

  • Since a radio wave is a electromagnetic type of wave, it propagates approximately at the same speed of light in vacuum, 3*10⁸ m/s.
  • As in any wave, there exist a fixed relationship between the propagation speed, the wavelength and the frequency of the radio wave, as follows:

       c = \lambda* f (1)

  • Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:

       \lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m  (2)

8 0
2 years ago
a care starting from rest has an acceleration 0.3 m/s square, calculate the velocity and distance travelled by this car after 2
Anna [14]

Answer:

Final velocity (v) = 36 m/s

Distance traveled (s) = 2,160 m

Explanation:

Given:

Initial velocity (u) = 0

Acceleration (a) = 0.3 m/s

Time travel (t) = 2 minutes = 120 seconds

Find:

Final velocity (v) = ?

Distance traveled (s) = ?

Computation:

v = u + at

v = 0 + 0.3(120)

v = 0.3(120)

v = 36 m/s

Final velocity (v) = 36 m/s

Distance traveled (s) = ut + (1/2)at²

Distance traveled (s) = (0.5)(0.3 × 120 × 120)

Distance traveled (s) = 2,160 m

3 0
3 years ago
What type of current is produced by a battery?
nadya68 [22]
The type of current produced by a battery is direct current. C.
8 0
3 years ago
Read 2 more answers
A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is
Anika [276]

Answer:

The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.

Explanation:

For 2 quantities A and B represented as

A\pm \Delta A and B\pm \Delta B

The sum is represented as

Sum=(A+B)\pm (\Delta A+\Delta B)

For the the values given to us the sum is calculated as

Sum=(2.9+3.9)\pm (0.1+0.2)

Sum=6.8\pm 0.3

Now the since the uncertainity inthe sum is \pm 0.3

The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity

Thus closest distance equals 6.8-0.3=6.5meters

3 0
3 years ago
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