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nirvana33 [79]
3 years ago
13

Find the electric field a distance z above a circular ring carrying a constant line charge. For extra credit you may derive the

electric field a distance z above a disk carrying a constant surface charge density. 3 Find the force between an electron and a proton in a typical hvdrogen atom.

Physics
1 answer:
lora16 [44]3 years ago
4 0

Answer:

F = 8.23 × 10⁻⁸N

Explanation:

F= kq²/r²

k= 9×10⁹Nm²/C²

q= 1.6×10⁻¹⁹C

r= 5.29×10⁻¹¹m

F= 9×10⁹× (1.6×10⁻¹⁹)²/  (5.29×10⁻¹¹)²

F = 8.23 × 10⁻⁸N

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Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having
igomit [66]

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) \frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

6 0
3 years ago
A car starts from rest and accelerated at a rate of 9m/s squared for 6.5 seconds. How much distance was covered by the car?
ki77a [65]

Answer:

58.5 meters

Explanation:

1. Find your formula. Use distance = speed x time for this problem

2. Plug in the given information. d (for distance) = 9m/s^2 * 6.5 s

3. Multiply number AND units. d = 58.5m

4. Check to make sure units & numbers make sense. In this case check that the answer is a lot bigger than what we stated with and that our units go with distance

5 0
2 years ago
I need help on this!
sergij07 [2.7K]
Try looking it up. that might help
8 0
2 years ago
A force of 25.0 N is applied 30° above the horizontal to a 5.0 kg box to move it across a smooth surface having coefficient of k
nordsb [41]

Answer:2.87

Explanation:

Given

mass of force=25 N

mass of box=5 kg

coefficient of kinetic friction \mu _k=0.2

Acceleration of box will be provided by its cos component and friction try to oppose it

Normal reaction=mg-F\sin 30

N=5\times 9.8 -25\sin 30 =36.5 N

friction force(f_r)=\mu _k N=0.2\times 36.5=7.3 N

net force in horizontal direction

F\cos 30 -f_r=ma

25\cos 30-7.3=5\times a

a=2.87 m/s^2

8 0
2 years ago
How does gravity maintain the the atmospheric layer around earth ?
HACTEHA [7]

Answer:

Our atmosphere is a mixture of gases that surround Earth. It is kept in place by the pull of Earth's gravity. If Earth was a much smaller planet, like Mercury or Pluto, its gravity would be to weak to hold a large atmosphere

Explanation:

5 0
2 years ago
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