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Nady [450]
1 year ago
5

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

Physics
1 answer:
coldgirl [10]1 year 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.

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irga5000 [103]
Sometimes arithmetic problems can be solved much more easily using the dimensional analysis approach. You focus on the units of the given information. Then, you manipulate them applying the laws of algebra where like units cancel, in order to end up with the unit of the unknown.

Given:
-50 nc/step
31 steps
Unknown: charge

Thus,
Charge = -50 nc/step * 31 steps =<em> -1550 nc</em>
7 0
1 year ago
A standard 1 kilogram weight is a cylinder 51.0 mm in height and 42.0 mm in diameter. Determine the density of the material
worty [1.4K]

Answer:

14160 kg/m^3

Explanation:

First of all, we need to find the volume of the cylinder.

The volume of the cylinder is given by:

V=\pi r^2 h

where:

r=\frac{d}{2}=\frac{42.0 mm}{2}=21.0 mm=0.021 m is the radius

h=51.0 mm=0.051 m is the height

Substituting, we find

V=\pi (0.021 m)^2 (0.051 m)=7.1 \cdot 10^{-5} m^3

And the density is given by

d=\frac{m}{V}

where m = 1 kg is the mass. Substituting, we find

d=\frac{1 kg}{7.1\cdot 10^{-5} m^3}=14,160 kg/m^3

3 0
2 years ago
Classify the following as alkali metals, alkaline earth metals, transition elements, or inner transitional elements: calcium, go
Digiron [165]
Alkali metals : sodium , potassium
alkaline earth : magnesium , calcium
the rest are transition elements... i don't know about "inner transition"
6 0
1 year ago
Read 2 more answers
A Body OF Volume 36cc Floats With 3/4 of its volume submerged in water . The density Of Body is
Radda [10]

Answer:

Density of body = 0.25g/cc

Explanation:

Given:

Volume submerged in water = 3/4

Find:

Density Of Body

Computation:

Density of body = fraction of body in liquid x density of water

Density of body = [1-3/4]1

Density of body = 0.25g/cc

8 0
1 year ago
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start
FrozenT [24]

Answer:

  • The total distance traveled is 28 inches.
  • The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector \hat{i} pointing in the west direction, the ant start at position

\vec{r}_0 = 16 \ inch \ \hat{i}

Then, moves to

\vec{r}_1 = 29 \ inch \ \hat{i}

so, the distance traveled here is

d_1 = |\vec{r}_1 - \vec{r}_0  | = | 29 \ inch   \ \hat{i} - 16 \ inch   \ \hat{i}  |

d_1 =  | 13 \ inch   \ \hat{i}  |

d_1 =  13 \ inch

after this, the ant travels to

\vec{r}_2 = 14 \ inch \ \hat{i}

so, the distance traveled here is

d_2 = |\vec{r}_2 - \vec{r}_1  | = | 14 \ inch   \ \hat{i} - 29 \ inch   \ \hat{i}  |

d_2 =  | - 15 \ inch   \ \hat{i}  |

d_2 =  15 \ inch

The total distance traveled will be:

d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch

The displacement is the final position vector minus the initial position vector:

\vec{D}=\vec{r}_2 - \vec{r}_1

\vec{D}= 14 \ inch   \ \hat{i} - 16 \ inch \ \hat{i}

\vec{D}= - 2 \ inch \ \hat{i}

This is 2 inches to the east.

6 0
2 years ago
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