Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
Answer:
n = 1,875
Explanation:
The speed of light in vacuum is constant (c) and in a material medium it is
v = d / t
The refractive index of a material is defined by
n = c / v
Let's look for the speed of light in the material, in general the length that light travels is known, this value is high, x = 1, when we place a block on the road, a small amount is lengthened by the length of the block, which in general is despised
These measurements are made on a digital oscilloscope that allows to stop the signals and measure their differences, that is, the zero is taken when the first ray arrives and the time for the second ray is measured,
v = d / t
v = 1 / 6.25 10⁻⁹
v = 1.6 10⁸ m / s
we calculate the refractive index
n = 3 10⁸ / 1.6 10⁸
n = 1,875
I believe your answer is TRUE!
Hope this helps!:)
<h2>Given that,</h2>
Mass of two bumper cars, m₁ = m₂ = 125 kg
Initial speed of car X is, u₁ = 10 m/s
Initial speed of car Z is, u₂ = -12 m/s
Final speed of car Z, v₂ = 10 m/s
We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :
v₁ is the final speed of car X.
So, car X will move with a velocity of -12 m/s.
