In standing wave no of loop depend on what?

A dog is 60 m away while moving at a constant velocity of 10 m/s towards you. Where is the dog after 4 seconds?

joja [24]

The dog would be 20 m away from you.

4 0

3 years ago

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What is the name of the force that keeps the planets and other celestial bodies orbiting around the sun

zhenek [66]

Gravity is the name of the force that keeps the planets in orbit

6 0

3 years ago

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Can someone solve this problem and explain to me how you got it

Zarrin [17]

1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

$F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}$

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

$F=k\frac{q_1 q_2}{r^2}$

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

$q_1' = 8q_1$

$q_2' = 8q_2$

Substituting into the equation, we find the new force:

$F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F$

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

After a surface interval my pressure group is D. I want to do a repetitive dive to 38 feet for 32 minutes. What will be my new p

ExtremeBDS [4]

Answer:

New pressure group is N

Explanation:

The Repetitive Dive table has to be checked, the residual nitrogen for the maximum depth of 38 feet as a "D" diver, then add that to the actual dive time to get the total nitrogen time. The total nitrogen time and the maximum depth of 38 feet would put you on a pressure group N.

5 0

3 years ago

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A net force of magnitude 36 N gives an object of mass m1 an acceleration of 6.0 m/s^2. The same net force gives m1 and another o

Jet001 [13]

This is a beautiful problem to test whether a student actually understands
Newton's 2nd law of motion . . . Force = (mass) x (acceleration).

That simple law is all you need to solve this problem, but you need to
use it a few times.

m₁ alone:
   Force = (mass) x (acceleration)

   36 N = ( m₁ ) x (6 m/s²)

   m₁ = (36 N) / (6 m/s²)

   m₁ = 6 kilograms .

m₁ and m₂ glued together:
   Force = (mass) x (acceleration)

   36 N = (6 kg + m₂) x (2 m/s²)

   6 kg + m₂ = (36 N) / (2 m/s²) = 18 kilograms

   m₂ = 12 kilograms .

m₂ alone:
   Force = (mass) x (acceleration)

   36 N = (12 kg) x (acceleration)

   Acceleration = (36 N) / (12 kg)

   Acceleration =   3 m/s²

3 0

4 years ago