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slava [35]
3 years ago
6

Please answer my question

Physics
2 answers:
hram777 [196]3 years ago
7 0

Answer:

Answer is (b) Mercury, venus and Mars.

Flura [38]3 years ago
3 0

Explanation:

i think b is correct!!

;-) :-) :-) :-)

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A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of 0.600c and 0.600c. What is the
vaieri [72.5K]

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

5 0
3 years ago
Arunner starts from rest and accelerates uniformly to a speed of 8.0 meters per
VladimirAG [237]

Answer:

2.0 m/s/s

Explanation:

The acceleration of an object is the rate of change of velocity of the object.

Mathematically, it is given by:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

Acceleration is a vector, so it has both a magnitude and a direction.

For the runner in this problem, we have:

u = 0 is the initial velocity (he starts from rest)

v = 8.0 m/s is the final velocity

t = 4.0 s is the time taken

Substituting, we find

a=\frac{8.0-0}{4.0}=2.0 m/s^2

3 0
3 years ago
Two balls, ball A and ball B, are dropped from the same height onto the same surface. If ball A rebounds to a higher height than
mamaluj [8]

Answer:

its b

Explanation:

3 0
3 years ago
To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that
noname [10]

Answer:

a. 2v₀/a   b. 2v₀/a  

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster

s' = 0t + 1/2at²

s' = 1/2at²

Since the distance moved by me and the dragster must be the same,

s = s'

v₀t. =  1/2at²

v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a  

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s =  v₀(2v₀/a)

= 2v₀/a  

4 0
3 years ago
39. A dog runs on a waxed floor at an initial speed of 2 m/s. It slides to a stop with an
Sedbober [7]

Answer:

Explanation:

Use the one-dimensional equation

v_f=v_0+at where vf is the final velocity of the dog, v0 is the initial velocity of the dog, a is the acceleration of the dog, and t is the time it takesto reach that final velocity. For us:

0 = 2 + -.43t and

-2 = -.43t so

t = 4.7 seconds

5 0
3 years ago
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