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dlinn [17]
3 years ago
9

The strength of the pull of gravity depends on _______. (1 point) speed and distance mass and distance mass and speed speed and

inertia
Physics
1 answer:
KIM [24]3 years ago
7 0
F_g=Gm_1m_2/r^2. mass and distance determines gravity.
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Table salt is blank mixed with other compunds
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Sodium mixed with chloride 
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Why is it that parents establish a positive relationship and positive communication with babies and young children
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In order to gain their trust, so that in any situation in their lifes like any worry their in, they can trust their parents for solutions in this wore some moment.
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3 years ago
Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t
yawa3891 [41]

Answer:

a) x=2*10^{9} m and t=8.35 s

b) t = γt', so it is 8.35 s.

Explanation:

a) The equation of Lorentz transformations is given by:

x=\gamma(x'+ut')  

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

  • x' = 0
  • t' = 5.00 s
  • u =0.800 c, c is the speed of light 3*10⁸ m/s

\gamma=\frac{1}{\sqrt{1-(u/c)^{2}}}

\gamma=\frac{1}{\sqrt{1-(0.800c/c)^{2}}}

\gamma=\frac{1}{\sqrt{1-(0.800)^{2}}}

\gamma=1.67

x=1.67(0+0.800c*5.00)

x=2*10^{9} m

Now, to find t we apply the same analysis:

t=\gamma(t'+\frac{ux'}{c^{2}})                        

but as x'=0 we just have:

t=\gamma(t')

t=1.67*5.00=8.35 s

b) Here, Mavis reads 5 s on her watch and Stanley measured the events at a time affected by the Lorentz factor, in other words t = γt', if we see it is the same a) part. So the time interval will be equal to 8.35 s.

I hope it helps you!

5 0
3 years ago
Read 2 more answers
A hydraulic lift raises a 4000 kg automobile when a 500N force is applied to the smaller piston. If the smaller piston has an ar
MariettaO [177]

Answer:

The cross-sectional area of the larger piston is 800 cm².

Explanation:

Given;

mass of the automobile, m = 4000 kg

force applied on the small piston, F₁ = 500 N

area of the smaller piston, A₁ = 10 cm²

load lifted by the larger piston, F₂ = 4000 x 10 = 40,000 N

Pressure experienced by each piston is given as;

\frac{F_1}{A_1} = \frac{F_2}{A_2}

Where;

A₂ is the cross-sectional area of the larger piston

\frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\A_2 = \frac{F_2A_1}{F_1} \\\\A_2 = \frac{40,000 \ \times\ 10}{500} \\\\A_2=800 \ cm^2

Therefore, the cross-sectional area of the larger piston is 800 cm².

8 0
3 years ago
What carries electrons from the power supply to a load?
Andru [333]
The answer is A! Hope it’s correct’
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