Answer:
I think the answer will be water ,sorry if ik wrong
First of all, you didn't tell us WHO measured the "10 years".
If it was the people on Earth, then 10 years passed according to them.
If it was 10 years on the space traveler's clock, then the clock in the
OTHER place, like on Earth, is subject to the relativistic 'time dilation'.
If the clocks are moving relative to each other, then the time interval measured
on either clock is equal to the interval measured on the other clock, divided by
√(1 - v²/c²) .
You said that v/c = 0.85 .
v²/c² = (0.85)² = 0.7225
1 - v²/c² = 1 - 0.7225 = 0.2775
√(1 - v²/c²) = √0.2775 = 0.5268
If one clock counts up 10 years, then the other one counts up
(10years) / 0.5268 = <em>18.983 years </em>
I believe that's the way to do this, and I'll gladly take your points,
but let me recommend that you get a second opinion before you
actually take off on your 10-year interstellar mission.
Hello there
the answer is
Let the initial position from where the ball is kickedat angle 45 deg, be A. The Horizontal range of the ball (AC) is V2 Sin2θ / g = 38.76 Meters
thank you
451539497.049 is the answer
Answer:
18301.4Kg
Explanation:
Step 1:
Data obtained from the question. This include the following:
Mass 1 (M1) = 0.512Kg
Mass 2 (M2) =..?
Distance apart (r) = 0.0250m
Force (F) = 0.001N
Gravitational force constant (G) = 6.67x10^-11Nm²/Kg²
Step 2:
Determination of the mass, M2 needed to create a force of 0.001N.
This can be obtained as follow:
F = GM1M2/r²
0.001 = 6.67x10^-11 x 0.512 xM2/0.025²
Cross multiply
6.67x10^-11 x 0.512 xM2 = 0.001x0.025²
Divide both side by 6.67x10^-11 x 0.512
M2 = (0.001x0.025²)/(6.67x10^-11x0.512)
M2 = 18301.4Kg
Therefore, a mass of 18301.4Kg is needed
