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Cloud [144]
3 years ago
14

Two objects with a mass of 10,000-kg each and a distance of 1 meter between them.

Physics
1 answer:
Fofino [41]3 years ago
4 0

Answer:

d) What is the force if we doubled both the masses AND we doubled the distance

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On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distanc
Arisa [49]

Answer:

Explanation:

The difference in time will be due to travel through atmosphere where speed of light slows down. If t be the thickness of atmosphere and c be the speed of light in space and μ be the refractive index of atmosphere difference in travel time will be as follows .

difference = \frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( 1-\mu  \right )

Now t = 40 x 10³m ,μ = 1.000293 , c = 3 x 10⁸.

difference =\frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( \mu -1  \right )\\

=\frac{ 2\times 40\times 10^3}{3\times10^3 }\left ( 1.000293-1 \right )\\

=7.81\times 10^{-3} s

4 0
3 years ago
A boat with a horizontal tow rope pulls a water skier. She skis off to the side, so the rope makes an angle of 12.0 ∘ with the f
Sladkaya [172]

Answer:

62.601kJ

Explanation:

We have a force with a horizontal component given by the angle of 15°. So wue can take this component to calculate the total work done, that is,

W_T = Fd(cos\theta)

Our values are,

d=320m

Angle = 12\°

Force=200N,

Substituting in Work equation,

W_T = 200(320)(cos12)

W_T = 62601.4J

W_T = 62.601kJ

4 0
3 years ago
Analyze how buffers allow you to eat acidic and basic foods without changing your blood pH.
blondinia [14]
Buffers neutralize the acid and the bases
7 0
3 years ago
_ is where there are no particles or very few particles that are spaced out very far apart. (example: outer space)
irina1246 [14]
The answer is d - Vacuum
4 0
3 years ago
7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del
GrogVix [38]

Answer:

 x =  0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

      sin 60 = T_{y} / T

      T_{y} = T sin 60

       cos 60 = Tₓ / T

      Tₓ = T cos 60

we apply the equation

       ∑ τ = 0

       -W L / 2 - w x + T_{y} L = 0

 

the length of the bar is L = 6m

           -Mg 6/2 - m g x + T sin 60 6 = 0

             x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

             x = (4676 - 5880) / 6860

             x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0
3 years ago
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