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Aleonysh [2.5K]
3 years ago
8

1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t

he electric force? The constant k = 9 x 109 N-m2/C2
Solution:
Physics
1 answer:
densk [106]3 years ago
3 0

Answer:

F = 36 N

Explanation:

Given that,

Charge, q₁ = +8 μC

Charge, q₂ = -5 μC

The distance between the charges, r = 10 cm = 0.1 m

We need to find the magnitude of the electrostatic force. The formula for the electrostatic force is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\F=36\ N

So, the magnitude of the electrostatic force is 36 N.

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A satellite omass1000 kg moves in a circular orbit of radius 8000 km round the earth,assumed to be a sphere of radius 6400 km. C
lubasha [3.4K]

Answer:

ΔE = 37.8 x 10^9 J

Explanation:

The energy required will increased the potential energy and increase the kinetic energy.

As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative

Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²

G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²

ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J

The centripetal force at orbit must be equal to the gravity force

mv²/R = mg'

v²/8.0e6 = 6.272

v² = (6.272(8.0e6)) = 50.2e6 m²/s²

The maximum velocity when resting on earth at the equator is about 460 m/s.

The change in kinetic energy is

ΔKE = ½m(vf² - vi²)(1000)

ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J

Total energy increase is

25e9 + 12.857e9 = 37.8e9 J

3 0
3 years ago
Please answer it is urgent <br>I will make u brainlist
adelina 88 [10]
For B, it is because water is a really good conductor of electricity, so the electrician will get shocked
6 0
3 years ago
A bumblebee
sattari [20]

Answer:

given -

initial velocity = 4.09 m/s

acceleration = 1.01 m/s²

distance = 23.4 m

time = ?

using second formula of motion,

s = ut + 1/2 at².

where, s = distance

u = initial velocity

t = time

a = acceleration

23.4 = 4.09(t) + 1/2(1.01)(t) ²

23.4 = 4.09t + 2.02t²

2.02t² + 4.09t - 23.4 = 0

solving the equation by using quadratic formula

Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :

a = 2.02

b = 4.09

c = -23.4

we get t=2.539 or t= -4.563

time cannot be negative so

t= 2.539 sec = 2.6 Sec is the answer

3 0
2 years ago
What is the speed over the ground of an airplane flying at 100 km/h relative to the air caught in a 100-km/h right-angle crosswi
Rudiy27

Answer:

141.42 km/h

Explanation:

Since this is a 100km/h right angle crosswind, the speed vector with respect to the ground is the vector speed of the airplane with respect to air + the vector speed of the cross-wind with respect to the ground

<100,0> + <0,100> = <100,100>

This vector has a magnitude of

v = \sqrt{v_1^2 + v_2^2} = \sqrt{100^2 + 100^2} = \sqrt{10000 + 10000} = \sqrt{20000} = 141.42  km/h

4 0
3 years ago
A ball is thrown straight up at an initial speed of 9.8m/s and, on returning to your hand, hits it moving downward at the same s
likoan [24]

Answer:

-9.8 m/s²

0 m/s

Explanation:

Average acceleration is the change in velocity over change in time.

a = Δv / Δt

a = (-9.8 m/s − 9.8 m/s) / 2.0 s

a = -9.8 m/s²

Average velocity is the change in position over change in time.

v = Δx / Δt

v = (0 m − 0 m) / 2.0 s

v = 0 m/s

3 0
3 years ago
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