All categories

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:

<em>The gravitational force between Royce and Earth would be doubled at 16 years.</em>

Explanation:

<em>"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".</em>

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.

You might be interested in

Magnetic fields within a sunspot can be as strong as 0.4T. (By comparison, the earth's magnetic field is about 1/10,000 as stron

WARRIOR [948]

Answer:

The speed of ejection is $2.06\times 10^{4}\ m/s$

Solution:

As per the question:

Magnetic field density, B = 0.4 T

Density of the material in the sunspot, $\rho = 3\times 10^{4}\ kg/m^{3}$

Now,

To calculate the speed of ejection of the material, v:

The magnetic field energy density is given by:

$U_{B} = \frac{B^{2}}{2\mu_{o}}$

This energy density equals the kinetic energy supplied by the field.

Thus

$KE = U_{B}$

$\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}$

where

m = mass of the sunspot in $1\ m^{3}$ = $3\times 10^{- 4}\ kg/m^{3}$

$v = \frac{B}{\sqrt{\mu_{o}m}}$

$v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s$

3 0

2 years ago

By examining a wave's pattern after it interacts with a barrier or gap, measurements can be made to better understand certain wa

rjkz [21]

Sample Response: How does diffraction occur?

8 0

2 years ago

Read 2 more answers

A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

malfutka [58]

Answer: A) $6.38(10)^{6} m$

Explanation:

The equation for the moment of inertia $I$ of a sphere is:

$I=\frac{2}{5}mr^{2}$ (1)

Where:

$I=9.74(10)^{37}kg m^{2}$ is the moment of inertia of the planet (assumed with the shape of a sphere)

$m=5.98(10)^{24}kg$ is the mass of the planet

$r$ is the radius of the planet

Isolating $r$ from (1):

$r=\sqrt{\frac{5I}{2m}}$ (2)

Solving:

$r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}}$ (3)

Finally:

$r=6381149.077m \approx 6.38(10)^{6} m$

Therefore, the correct option is A.

4 0

1 year ago

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hourmiles/hour

slavikrds [6]

Answer:

The acceleration of the cheetahs is 10.1 m/s²

Explanation:

Hi there!

The equation of velocity of an object moving along a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time t.

v0 = initial velocity.

a = acceleration.

t = time

We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.

Let's convert mi/h into m/s:

50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s

Then, using the equation:

v = v0 + a · t

22.4 m/s = 0 m/s + a · 2.22 s

Solving for a:

22.4 m/s / 2.22 s = a

a = 10.1 m/s²

The acceleration of the cheetahs is 10.1 m/s²

5 0

2 years ago

A submarine completed a 450 km training with an average speed of 50 km/h. For the first 180 km, it travelled at an average speed

Kryger [21]

Answer:

45km/hr

Explanation:

Total distance=450km

Total speed=50km/hr

Total time= distance/speed

=450/50

=9hrs

distance a=180km

speed a=60km/hr

Time a=180/60

=3hrs

Distance b=450-180=270km

Speed b=?

Time b=270/speed b

Total time=time a + time b

9=3+(270/speed b)

270/speed b =9-3

270/speed b =6

6*speed b =270

Speed b=270/6

Speed b=45km/hr

4 0

2 years ago