All categories

2 years ago

3. If the radius of Earth decreased, with no change in mass, your weight would

Physics

1 answer:

m_a_m_a [10]2 years ago

6 0

Answer:

The weight of the person would increase

Explanation:

The Universal Law of Gravitation gives the magnitude of the force between the masses of two objects (m1 and m2) separated a given distance "d" as:

$F_g=G\,\frac{m_1*m_2}{d^2}$

where G is the universal gravitational constant.

Our weight on Earth is this force between the Earth (of mass M) and ourselves (our mass m) at a distance that is the Earth's radius R:

$Weight=G\frac{M*m}{R^2}$

Now, if we keep all the values equal (mass of the Earth M and our mass m) except for the distance between the center of the Earth and our center of gravity (the radius of the Earth), we are going to have now a smaller radius (r) in the formula above:

$Weight=G\frac{M*m}{r^2}$

and dividing by a smaller number (r is smaller than R), will render a larger quotient. This means that the actual force (weight) will become larger, so the weight would clearly increase.

You might be interested in

RACTIC PTUDIES

Amanda [17]

Answer:

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

Explanation:

The distance of the book before the lamp is moved, $d_{b} = 30 cm$

The distance of the book after the lamp is moved, $d_{a} = 90 cm$

Illumination can be given by the formula, $E = \frac{P}{4 \pi d^{2} }$

Illumination before the lamp is moved, $E_{b} = \frac{P}{4 \pi d_{b} ^{2} }$

Illumination after the lamp is moved, $E_{a} = \frac{P}{4 \pi d_{a} ^{2} }$

$\frac{E_{a}}{E_{b}} } = \frac{\frac{P}{4 \pi d_{a} ^{2} } }{\frac{P}{4 \pi d_{b} ^{2} } }$

$\frac{E_{a} }{E_{b} } = \frac{d_{b} ^{2} }{d_{a} ^{2}} \\\frac{E_{a} }{E_{b} } = \frac{30^{2} }{90 ^{2}}\\\frac{E_{a} }{E_{b} } =\frac{900 }{8100}\\\frac{E_{a} }{E_{b} } =\frac{1 }{9}$

$E_{b} = 9E_{a}$

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

6 0

2 years ago

You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a batte

VikaD [51]

Answer:

$R_3 < R_1 < R_2$

Explanation:

The resistance of a wire is given by:

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

1) The first wire has length L and cross-sectional area A. So, its resistance is:

$R_1=\frac{\rho L}{A}$

2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is

$R_2=\frac{2\rho L}{A}$

3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is

$R_3=\frac{\rho L}{2A}$

By comparing the three expressions, we find

$R_3 < R_1 < R_2$

So, this is the ranking of the wire from most current (least resistance) to least current (most resistance).

5 0

2 years ago

1. A positive electric charge in a region of changing electric potential will: A. move in the direction of decreasing potential

julia-pushkina [17]

Answer:

The correct option is (B).

Explanation:

The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. The electric potential due to a point charge is given by :

$V=\dfrac{kQ}{r}$

Where

k is the electrostatic constant

Q is the electric charge

r is the distance from the charge

So, a positive electric charge in a region of changing electric potential will move in the direction of increasing potential.

5 0

2 years ago

the passengers in a roller coaster feel 50 heavier than their true weight as the car goes through a dip with a 30m radius of cur

NeTakaya

Answer:

v = 12.12 m/s

Explanation:

Given that,

Radius of the curvature, r = 30 m

To find,

The car's speed at the bottom of the dip.

Solution,

Let mg is the true weight of the passenger. When it is moving in the circular path, the centripetal force act on it. It is given by :

$F=\dfrac{mv^2}{r}$

The normal reaction of the passenger is given by :

$N=mg-50\%mg$

N = 1.5 mg

Let v is the car's speed at the bottom of the dip. It can be calculated as:

$1.5\ mg-mg=\dfrac{mv^2}{r}$

$v=\sqrt{0.5gr}$

$v=\sqrt{0.5\times 9.8\times 30}$

v = 12.12 m/s

So, the speed of the car at the bottom of the dip is 12.12 m/s. Hence, this is the required solution.

7 0

2 years ago

How does the body maintain homeostasis? Give two or more examples of how this is accomplished. Provide supporting details. Be de

Minchanka [31]

Explanation:

-body temp: when cold, erector muscle contract, hair raise, vasoconstriction occurs, sweat production reduces, muscle shiver to enhance metabolism, and thyroid gland secret more thyroxine to enhance metabolism

-water potential balance: when we are thirsty, pituitary gland secret more ADH, kidney tubules become more permeable to water, so more water re-absorbed, less urine produced, retaining more water.

8 0

2 years ago