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elena-14-01-66 [18.8K]
3 years ago
9

Match the description with the type of front

Physics
1 answer:
Inga [223]3 years ago
4 0

Answer: Do you need help with how to get the photograph?

Explanation:It's in the comments if you want to known have a good day!

:)

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The magnitude of the momentum of an object is the product of its mass m and speed v. If m = 3 kg and v = 1.5 m/s, what is the ma
Oliga [24]

Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

p = m v

Mass, m = 3 kg

Velocity, v = 1.5 m/s

So, momentum of this object is given by :

p=3\ kg\times 1.5\ m/s

p = 4.5 kg-m/s

or

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

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3 years ago
Technology is the application of science to make products that are useful to people. True or False
alekssr [168]
False false false false
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3 years ago
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How do I use a metric conversions chart
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4 years ago
Lists the 4 states of matter from most amount of thermal energy to least amount of thermal energy?
Cerrena [4.2K]

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3 years ago
Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?
sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that  describes the electrostatic force between two charged particles. If the particles have charges q_1 and q_2, are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

|F|=k \cfrac{q_1 q_2}{r^2}

Thus if we decrease the distance by half we have

r_1 =\cfrac r2

So we get

|F|=k \cfrac{q_1 q_2}{r_1^2}

Replacing we get

|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}

We can then multiply both numerator and denominator by 4 to get

|F|=k \cfrac{4q_1 q_2}{r^2}

So we have

|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

q_{2new} =2q_2

So we get

|F|=k \cfrac{q_1 q_{2new}}{r_1^2}

which give us

|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)

Thus doubling one of the charges doubles the force.

So the answer is A.

8 0
3 years ago
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