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Goryan [66]
3 years ago
10

When point charges q1 = +7.9 μC and q2 = +6.0 μC are brought near each other, each experiences a repulsive force of magnitude 0.

75 N. Determine the distance between the charges.
Physics
1 answer:
cricket20 [7]3 years ago
5 0

Answer:0.754m

Explanation:

F=kq1q2/r^2

R^2= kq1q2/f

R^2= 9*10^9*7.9*10^-6*6.0*10^-6/0.75

R^2= 9*7.9*6*10^-3/0.75

R^2=0.4266/0.75

R^2=0.5688

R=√0.5688

R=0.754m

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ratelena [41]

Answer:

Explanation:

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We are surrounded by mixtures. The air is a mixture of gases . The oceans are a mixture of (mainly) water and salt. The solid earth is mostly rock, which is a mixture of different minerals.

Natural resources are substances we need and use, which occur naturally. Some come from living things, (example) cotton other are non-living (example)  sand.

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5 0
2 years ago
Today chemistry is based on the law ofwhat is the matter ​
melisa1 [442]

Answer:

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Explanation:

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5 0
3 years ago
When the distance between two stars decreases by one-third, the force between them
pashok25 [27]

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

r'=\frac{2}{3}r

So the new force will be

F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F

So, the force will be 2.25 times the previous value.

4 0
3 years ago
A bus starts to move from rest. If its velocity becomes 90 km/hr after 8 seconds, calculate its acceleration.
a_sh-v [17]

Answer:

The acceleration is 11.25 km/hr

Explanation:

Divide the velocity over time to find the acceleration. The bus is accelerating 11.25 km/hr per second in the range of 8 seconds

7 0
3 years ago
g Two radiation modes (one at the center frequency lIo and the other at lIO+?lI) are excited with 1000 photons each. Determine t
Schach [20]

Answer:

a) P=0.25x10^-7

b) R=B*N2*E

c) N=1.33x10^9 photons

Explanation:

a) the spontaneous emission rate is equal to:

1/tsp=1/3 ms

the stimulated emission rate is equal to:

pst=(N*C*o(v))/V

where

o(v)=((λ^2*A)/(8*π*u^2))g(v)

g(v)=2/(π*deltav)

o(v)=(λ^2)/(4*π*tp*deltav)

Replacing values:

o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2

the probability is equal to:

P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7

b) the rate of decay is equal to:

R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system

c) the number of photons is equal to:

N=(1/tsp)*(V/C*o)

Replacing:

N=100/(3*3x10^10*8.3x10^-19)

N=1.33x10^9 photons

7 0
3 years ago
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