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Elena-2011 [213]
3 years ago
15

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

between them doubles?
Physics
1 answer:
antoniya [11.8K]3 years ago
7 0
The equation for gravitational force is:

F = GMm/r^2

Where G is a constant, M is the mass of the larger object, m is the mass of the other object, and r is the distance between the two objects.

So:

1600 = GMm/r^2

Now we can find the force when the distance doubles:

F = GMm/(2r)^2

GMm will remain the same since G is a constant and the masses didn't change, so the only difference between this and the first calculation is the denominator.

By simplifying the denominator, we find:

F = GMm/(4r^2)

Now we can see that the only difference between this and the equation given the original distance is that we are dividing by 4. This means that the force will be one fourth of the original force:

1600/4 = 400

The answer is 400.
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Gravity will always affect the object the most with the greater mass. So a stapler has a greater mass than all the other objects given. I would then assume that the correct answer is D.
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Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo
Elena-2011 [213]

a) 3.14 \cdot 10^{-4} s

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t] (1)

The generic equation that describes a wave is

y(t)=A sin (\frac{2\pi}{T} t) (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}

Therefore, the period is

T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when sin(...)=1, and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

\lambda=2L

where

\lambda is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

\lambda =2(5.00)=10.0 m

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]

And by comparing it with the general equation of a wave:

y(t)=A sin (\frac{2\pi}{T} t)

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

sin(\frac{2\pi}{T}t)=1

which means that

y(t)=A

And therefore in this case,

y=0.500 cm

So, this is the displacement.

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3 years ago
While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction b
tresset_1 [31]

Answer:

64.1°

Explanation:

Coefficient of friction = Horizontally resolved pushing force divided by Moving force

Coefficient of friction is 0.437

Resolving pushing force to horizontal = Moving force * Cos ∅

Therefore, 0.437 = Cos ∅

and ∅ = Arch Cos 0.437 = 64.1°

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A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an angle of 35°
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<em>Answer</em>


Force = 170 N



<em>Explanation</em>

First find the distance (d) travelled by the bulldozer.


Sin 35 = 15/d

d = 15/(sin 35)

= 26.15m


Now;

work done = force × distance.


4500 J = force × 26.15


dividing both sides by 26.15,


Force = 4500/26.15

= 172.07 N


Answer to two significant figures = 170 N

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What are the horizontal and vertical velocities of a stunt bike that leaves a ramp at 100 km/hr and at an angle of 35 degrees?
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Horizontal velocity: 81.9 km/h

Vertical velocity: 57.4 km/h

Explanation:

We can solve this problem by resolving the velocity vector into its component along the horizontal and vertical direction.

The horizontal velocity of the stunt bike is given by:

v_x = v cos \theta

where

v = 100 km/h is the magnitude of the velocity

\theta=35^{\circ} is the angle of projection

Substituting, we find

v_x = (100)(cos 35^{\circ})=81.9 km/h

The vertical velocity instead is given by

v_y = v sin \theta

where

v=100 km/h

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Substituting,

v_y = (100)(sin 35^{\circ})=57.4 km/h

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

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