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3 years ago

When a ball bounces back when you throw it at a wall, what occurs?

Physics

2 answers:

NISA [10]3 years ago

5 0

Answer:

a (refraction) i think

Explanation:

Ray Of Light [21]3 years ago

4 0

I would have to think a. Nothing else makes since

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What is the relation between acceleration due to gravity and radius of the earth?

emmasim [6.3K]

Answer:

As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

Formula: g = GM/r2

Dimensional Formula: M0L1T-2

Values of g in SI: 9.806 ms-2

Explanation:

Please Mark me brainliest

8 0

3 years ago

Read 2 more answers

A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

$E_{1}+E_{3}-E_{2}=0$

We know that the electric field is:

$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
q is the charge
r is the distance from the charge to the point

So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

$\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0$

$r_{3}=0.0743\:$

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

3 0

3 years ago

1. The sun radiates light from space to heat a swimming pool on Earth.

scoundrel [369]

Answer:

c. Light energy to thermal energy

Explanation:

The energy from the sun comes in the form of light energy but is converted to thermal energy.

7 0

2 years ago

Describe what happens to pressure as the force exerted on a given area increases.

Goryan [66]

That's the definition of pressure ... force on a given area.

So when that force increases, it's an increase in pressure.

4 0

3 years ago

The magnetic force on a straight wire 0.69 m long is 1.5 x 10-3 N. The current in the wire is 16.9 A. What is the magnitude of t

VikaD [51]

Answer:

Magnitude of magnetic field is 1.29 x 10⁻⁴ T

Explanation:

Given :

Current flowing through the wire, I = 16.9 A

Length of wire. L = 0.69 m

Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N

Consider B be the applied magnetic field.

The relation to determine the magnetic force experienced by current carrying wire is:

F = ILBsinθ

Here θ is the angle between magnetic field and current carrying wire.

According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:

F = ILB

$B=\frac{F}{IL}$

Substitute the suitable values in the above equation.

$B=\frac{1.5\times10^{-3} }{16.9\times0.69}$

B = 1.29 x 10⁻⁴ T

7 0

4 years ago