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3 years ago

Explain the relationships between gravity, mass, and distance.

Physics

2 answers:

Mazyrski [523]3 years ago

7 0

Answer:

As the mass of an object increases, its gravitational force increases.

As an object's distance to other objects increases, its gravitational force on those objects increases.

Explanation:

The gravitational force of one object on another is calculated with the equation

F = (G*m1*m2)/(r²),

where G is the gravitational constant,

M1 and M2 are the masses of the two objects, and

r is the distance between them

We can see that the force has a direct relationship with both of the mass values, and an inverse square relationship with the distance between them.

Hope this helped!

dexar [7]3 years ago

7 0

Simpler way to put it:

Gravity has a direct relationship with mass. An increase in mass results in an increase in gravitational force. Gravity also has an inverse relationship with distance. An increase in distance results in a decrease in gravitational force.

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What is the approximate efficiency of the engine?

Elodia [21]

Answer:91% I think

Explanation:

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3 years ago

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How do I do this physics problem about potential energy and kinetic energy?

larisa86 [58]

Ok i apologise for the messy working but I'll try and explain my attempt at logic

Also note i ignore any air resistance for this.

First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.

Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.

I also noted the "initial" and "final" height of the swing with hi and hf respectively.

So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.

So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).

Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.

The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).

I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx

4 0

3 years ago

A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal

zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity $u_{y}=0$

The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , $u_{y}=0$

Substitute these values in the rule

→ -60 = 0 + $\frac{1}{2}$ (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* The ball is in the air for 3.5 seconds 

The initial velocity is the horizontal component $u_{x}$

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = $u_{x}$ (3.5)

Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

→ Its direction $tan^{-1}\frac{-34.3}{28.6}=-50.18$

The negative sign means the direction is below the horizontal

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

7 0

3 years ago

The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t

KatRina [158]

Answer:

T₂ = 123.9 N, θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

T₁ₓ - $T_{2x}$ = 0

T_{2x}= T₁ₓ

Y axis y

T_{1y} + T_{2y} - 200N = 0

T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

sin 60 = T_{1y} / T₁

cos 60 = t₁ₓ / T₁

T_{1y} = T₁ sin 60

T1x = T₁ cos 60

T_{1y}y = 100 sin 60 = 86.6 N

T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

T_{2y} = T₂ sin θ

T₂ₓ = T₂ cos θ

we substitute

T₂ sin θ= 200 - 86.6 = 113.4

T₂ cos θ = 50 (1)

to solve the system we divide the two equations

tan θ = 113.4 / 50

θ = tan⁻¹ 2,268

θ = 66.2º

we caption in equation 1

T₂ cos 66.2 = 50

T₂ = 50 / cos 66.2

T₂ = 123.9 N

8 0

3 years ago

how much does a bookshelf weigh if the movers are pushing it at a speed of 10 m/s^2 by applying 100 N force

Delicious77 [7]

Answer:

10 kg

Explanation:

Assuming a frictionless surface, then force F=ma where F is the applied force, m is the mass and a is acceleration. Making m the subject of the formula then $m=\frac {F}{a}$

Substituting 100 N for the applied force F and 10 m/s^2 for acceleration a then the value of m will be $m=\frac {100}{10}=10\ kgs$

Therefore, in terms of kilograms, the bookshelf weighs 10 Kg

7 0

3 years ago