Weather radar is used to locate precipitation<span>, calculate its motion, and estimate its type rain, snow, </span>hail etc
so, d.measuring temperature <span>is not a use for a weather radar
</span>effect of global warming-<span>heavier rainfall and flooding</span>

Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
Elastic potential energy.
When you stretch a rubber band it has the "potential" to do work, to fly in a given direction. In doing so it changes it's elastic potential energy to kinetic energy.
Answer:
Explanation:
Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg
Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²
For change in period of rotation we shall apply conservation of angular momentum law
I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .
I₁ / I₂ = ω₂ / ω₁
I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .
T₂ / T₁ = I₂ / I₁
(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²
1 + 5 / 3 x 2.3 x 10¹⁹ / M
= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴
= 1 + .0000064
T₂ = 24 (1 + .0000064)
= 24 hours + .55 s
change in length of the day = .55 s .
Answer:
ΔF=125.22 %
Explanation:
We know that drag force on the car given as

=Drag coefficient
A=Projected area
v=Velocity
ρ=Density
All other quantity are constant so we can say that drag force and velocity can be given as

Now by putting the values



Percentage Change in the drag force



ΔF=125.22 %
Therefore force will increase by 125.22 %.