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3 years ago

By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)

Physics

1 answer:

Fofino [41]3 years ago

6 0

Gravity on the surface of sun is given as

$g = \frac{GM}{R^2}$

here we know that

$M = 1.98 \times 10^{30} kg$

$R = 6.95 \times 10^8 m$

now we will have

$g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}$

$g = 273.4 m/s^2$

now we need to find the ratio of weight on surface of sun and on surface of Earth

$R = \frac{mg_{sun}}{mg_{earth}}$

$R = \frac{273.4}{9.8} = 28 times$

so weight will increase by 28 times

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Two identical objects (A, B) travel circles of the same radius, but object A completes three times as many rotations as object B

malfutka [58]

Answer:

a) One-ninth the force acting on object A.

Explanation:

First, we derive an expression for the centripetal force acting on both objects.

For object A, centripetal force is:

$F_A = \frac{m{v_A}^2}{r}$

For object B, centripetal force is:

$F_B = \frac{m{v_B}^2}{r}$

We are given that they have the same mass and they move in circles of the same radius.

If object A completes three times as many rotations as object B, then, object must have 3 times the speed of object B.

Hence:

${v_A} = 3*{v_B}$

Therefore, $F_A$ becomes:

$F_A = \frac{m({3*v_B}^{2} )}{r}\\\\\\F_A = \frac{9m{v_B}^{2}}{r}$

$F_A = 9F_B$

=> $F_B = \frac{1}{9} F_A$

Therefore, the net centripetal force acting on object B is one-ninth of the force acting on object A.

4 0

3 years ago

A force of 10 N causes a spring to extend by 20 mm. Find: a) the spring constant of the spring in N/m

Mars2501 [29]

Answer:

formula used K=F/∆l

∆l is the elongation of the spring

F=10N
∆l=20mm===> 0.02m
K=10N divided 0.02m= 500N/m

6 0

3 years ago

A city located on the coast of North America has warmer winters and cooler summers than a city at the same elevation and latitud

Vesna [10]

I don't like any of those choices. But if you absolutely have to pick your answer
from this list, then it has to be 'D'.

The ocean is an enormous storage vessel for heat. It gets heat from the air in
the Summer ... which somewhat cools places near the coast ... and it releases
heat into the air during the Winter ... which warms places near the coast.

So I guess it's true that ocean surfaces change temperature more slowly than
land surfaces do, and they influence the land nearby in the process. But this
ignores the reason for the slow changes in ocean surface temperature. It's a
lot like saying that the loud noise produced by a race car is the result of the
car's ability to appear in a far different location after a short time.

5 0

3 years ago

vehicles equipped with disc brakes incorporate a mechanically operated drum-style parking brake in the center of the rear disc b

umka21 [38]

Vehicles equipped with disc brakes incorporate a mechanically operated drum-style parking brake in the center of the rear disc brake rotors, commonly called a top hat parking brake.

<h3>What is top hat parking brake?</h3>

The top hat drum style of parking brake system is called the top hat rotors because of how big the middle part is compared to regular rotors.

In this type of parking brake system, the vehicle is equipped with disc brakes which incorporate a mechanically operated drum-style parking brake in the center of the rear disc brake rotors.

Thus, vehicles equipped with disc brakes incorporate a mechanically operated drum-style parking brake in the center of the rear disc brake rotors, commonly called a top hat parking brake.

Learn more about top hat parking brake here: brainly.com/question/21854030

4 0

2 years ago

Pleaseeeee help me with b, c, and d. There are no angles.

taurus [48]

Answer:

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, $F_{NET}$ = F - $F_{Friction}$ = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

$F_{Friction}$ = The force of friction

∴ $F_{NET}$ = m × a = m × 0 m/s² = 0 N

Therefore;

$F_{NET}$ = F - $F_{Friction}$ = 0 N

From which we have;

F = $F_{Friction}$ = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = $F_{Friction}$ × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, $d_p$ = 0 m

∴ The work done by the normal direction = N × $d_p$ = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, $d_{wieght}$ = 0 m

Therefore, the work done by the weight = W × $d_{wieght}$ = W × 0 m = 0 J

The work done by the weight = 0 J

7 0

3 years ago