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2 years ago

When Royce was 10 years old, he had a mass of

Physics

1 answer:

coldgirl [10]2 years ago

6 0

Answer:

The gravitational force between Royce and Earth would be doubled at 16 years.

Explanation:

"Newton's law of universal gravitation states that gravitation force between two masses is proportional to the magnitude of their masses and inverse-squared of their distance".

Royce Scenario

At the age of 10 Royce's mass = 30kg

At the age of 16 Royce's mass = 60kg

From Newton's law of universal gravitation, an Increase in the mass of a body would amount to a corresponding increase in the gravitational force.

In the case of Royce, the mass double between the age of 10 and 16, so there would be an increase of the gravitation force by double.

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An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the

Marina86 [1]

Answer:

$2.88*10^{8} m/s$

Explanation:

The speed of light is given by

$c=\frac {2d}{t}$ and $t=\frac {\theta}{\omega}$ hence

$c=\omega\frac {2d}{\theta}$

Speed of light is given by

$c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s$

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2 years ago

Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

For more information on this visit

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2 years ago

A particle at 9 AM is moving towards the east at 4 ms At 12 noon, it changes its velocity and starts moving towards the north un

nexus9112 [7]

Answer:

Explanation:

Acceleration is the time rate of change of velocity.

Acceleration and velocity are vectors

If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.

There are 3 hours or 10800 seconds between 10 AM and 1 PM

a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²

There are 14400 seconds between 10 AM and 2 PM

The velocity changes are still the same

a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²

7 0

2 years ago

There is an electromagnetic wave traveling in the -z direction in a standard right-handed coordinate system. What is the directi

wlad13 [49]

Answer: The direction of the electric field, E→, is pointed in the +y direction.

Explanation:

One can use the right hand rule to illustrate the direction of travel of an electromagnetic and thereby get the directions of the electric field, magnetic field and direction of travel of the wave.

The right hand rule states that the direction of the thumb indicate the direction of travel of the electromagnetic wave (in this case the -z direction) and the curling of the fingers point in the direction of the magnetic field B→ (in this case the +x direction), therefore, the electric field direction E→ is in the direction of the fingers which would be pointed towards the +y direction.

6 0

2 years ago

A box slides down a frictionless plane inclined at an angle θ ¸ above the horizontal. The gravitational force on the box is dire

DedPeter [7]

<h2>Answer: at an angle $\theta$ below the inclined plane. </h2>

If we draw the Free Body Diagram for this situation (figure attached), taking into account only the gravity force in this case, we will see the weight $W$ of the block, which is directly proportional to the gravity acceleration $g$ :