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Vadim26 [7]
4 years ago
7

Help Please!!

Physics
1 answer:
Setler [38]4 years ago
3 0
A is the answer to the question
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According to Newton's law of universal gravitation, which statement is true?
valkas [14]

Answer:

Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

6 0
3 years ago
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It takes a photon 8 minutes and 25 seconds ro reach earth. What is the distance (labeled 1 au) in meter between the sun and eart
tangare [24]

Answer:

x=0.017 AU  

Explanation:

We can use the equation of speed in terms of distance. We know that the speed of light is constant value so we will have:

v=\frac{x}{t}

x=v*t

x=3*10^{8}*8.417  

x=2.53*10^{9}m  

Now we know that 1 AU = 1.496*10^{11}m

x=0.017 AU  

I hope it helps you!

3 0
3 years ago
What are the two forces that keep a pendulum swinging?
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The force of gravity is the only force that keeps a pendulum in motion. both the force increases the speed of the pendulum on the downswing and decreases it's speed on its upswing.
3 0
3 years ago
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Did changing the angle of the incline affect the following parameters?
MArishka [77]
Increasing the level of an incline:
Increases final velocity
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Increases the initial potential energy
Increases the final kinetic energy
4 0
4 years ago
A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a
stira [4]

Answer:

Explanation:

Given

acceleration is given by

a=-g-c_Dv^2

where \ddot{y}=a

\dot{y}=v

Also acceleration is given by

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

ds=\frac{v}{a}dv

\int ds=\int \frac{v}{-g-0.001v^2}dv

\Rightarrow Let -g-0.001v^2=t

-0.001\times 2vdv=dt

vdv=-\frac{dt}{0.002}

at\ v_0=50\ m/s,\ t=-g-0.001(50)^2

t=-g-2.5

at v=0,\ t=-g

\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}

\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}

s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}

s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})

s=113.608\ m

when air drag is neglected maximum height reached is

h=\frac{v_0^2}{2g}

h=\frac{50^2}{2\times 9.8}

h=127.55\ m

3 0
3 years ago
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