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navik [9.2K]
3 years ago
8

Please i need answer for part d

Physics
1 answer:
matrenka [14]3 years ago
4 0
It is c that’s what it is that’s the answer
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The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate
butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * 10^{8}km

speed of light = 3* 10^{8}m/s

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 *10^{8}*1000m

distance between earth and sun = 1.5 * 10^{11}m

speed = distance /time

time = distance / speed

time = \frac{1.5*10^{11} }{3*10^{8} }

= 0.5*10^{3}

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

3 0
3 years ago
Which of the following statements are true about magnets?
timama [110]
The correct answer is 3
4 0
3 years ago
A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
3 years ago
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