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ElenaW [278]
3 years ago
7

Plzzzzzz help :(((((((((((

Physics
2 answers:
OlgaM077 [116]3 years ago
7 0

Answer:

4. a) F

b) E

c) B

d) D

Hope it helps

RoseWind [281]3 years ago
6 0
F E B D, I think this is right...
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A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 10 nC is located at a distance r = 10
jok3333 [9.3K]

Answer:

q₁  = -2.92 nC

Explanation:

Given;

first point charge, q₁ = ?

second point charge, q₂ = 10 nC

net flux through the surface of the sphere, Φ =  800 N.m²/C

According to Gauss’s law, the flux through any closed surface (Gaussian surface), is equal to the net charge enclosed divided by the permittivity of free space.

\phi = \frac{q_{enc.}}{\epsilon_o}

where;

Φ is net flux

q_{enc.} net charge enclosed

ε₀ is permittivity of free space.

q_{enc.} = Φε₀

       = 800 x 8.85 x 10⁻¹²

       = 7.08 x 10⁻⁹ C

q_{enc.} = 7.08 nC

q₁ + q₂ = q_{enc.}

q₁ = q_{enc.} - q₂

q₁  = 7.08nC -  10 nC

q₁  = -2.92 nC

4 0
3 years ago
Mike and Mitchell decide to have a foot race. They mark off a stretch of 100 yards, and recruit Cindy to work the stopwatch.
kati45 [8]
D)The boys had times that matched exactly
3 0
3 years ago
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Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 m
Savatey [412]

Answer:

b) 0.5 N

Explanation:

From coulomb's law,

F = kq'q/r².................... Equation 1

Where F =force of repulsion between the charges, q' = first charge, q = second charge, r = distance between the charges, k = proportionality constant.

q'q = Fr²/k........................... Equation 2

Given: F = 2 N, r = 1 m, k = 9.0×10⁹ Nm²/C²

Substituting into equation 2

q'q = 2(1)²/(9.0×10⁹)

q'q =  2/9.0×10⁹ C².

If the distance between the charges is increased to 2 meters,

r = 2 m, q'q = 2/9.0×10⁹ C².

Substitute into equation 1

F = 9.0×10⁹(2/9.0×10⁹)/2²

F = 2/4

F = 1/2 = 0.5 N.

The right option is b) 0.5 N

5 0
4 years ago
What is not true ozone?
Gnoma [55]
The second answer choice is not true (ozone is concentrated in the stratosphere, not the exosphere)
3 0
3 years ago
The 10-lb block A attains a velocity of 2ft/s in 5 seconds, starting from rest. Determine the tension in the cord and the coeffi
Nezavi [6.7K]

Answer:

Explanation:

Let T be the tension in the cord.

Impulse by cord = change in momentum of block A .

T x 5s = 10 ( 2 -0) = 20

T = 4 poundal .

acceleration of block B = 2 / 5 = 0.4 m /s²

Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .

= 8 ( 32 + .4 ) = 259.2 poundal

Frictional force on block A = 259.2 - 4 = 255.2 poundal

μ x 10 x 32 = 255.2

320μ = 255.2

μ =0 .8 .

8 0
3 years ago
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