Answer:
Explanation:
3.4 m/s due North, -1.1 m/s due East
Answer:
V' = 0.84 m/s
Explanation:
given,
Linear speed of the ball, v = 2.85 m/s
rise of the ball, h = 0.53 m
Linear speed of the ball, v' = ?
rotation kinetic energy of the ball

I of the moment of inertia of the sphere

v = R ω
using conservation of energy


Applying conservation of energy
Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy



V'² = 0.7025
V' = 0.84 m/s
the linear speed of the ball at the top of ramp is equal to 0.84 m/s
A mechanical wave<span> is a </span>wave<span> that is an oscillation of </span>matter<span>, and therefore transfers energy through a </span>medium.[1]<span> While waves can move over long distances, the movement of the </span>medium of transmission<span>—the material—is limited. Therefore, oscillating material does not move far from its initial equilibrium position. Mechanical waves transport energy. This energy propagates in the same direction as the wave. Any kind of wave (mechanical or electromagnetic) has a certain energy. Mechanical waves can be produced only in media which possess elasticity and inertia.</span>
Answer:
(a) the electrical power generated for still summer day is 1013.032 W
(b)the electrical power generated for a breezy winter day is 1270.763 W
Explanation:
Given;
Area of panel = 2 m × 4 m, = 8m²
solar flux GS = 700 W/m²
absorptivity of the panel, αS = 0.83
efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp
panel emissivity , ε = 0.90
Apply energy balance equation to determine he electrical power generated;
transferred energy + generated energy = 0
(radiation + convection) + generated energy = 0
![[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0](https://tex.z-dn.net/?f=%5B%5Calpha_sG_s-%5Cepsilon%20%5Calpha%28T_p%5E4-T_s%5E4%29%5D-h%28T_p-T_%5Cinfty%29%20-%20%5Ceta%20%5Calpha_s%20G_s%20%3D%200)
![[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s](https://tex.z-dn.net/?f=%5B%5Calpha_sG_s-%5Cepsilon%20%5Calpha%28T_p%5E4-T_s%5E4%29%5D-h%28T_p-T_%5Cinfty%29%20-%20%280.553-0.001T_p%29%5Calpha_s%20G_s)
(a) the electrical power generated for still summer day

![[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k](https://tex.z-dn.net/?f=%5B0.83%2A700-0.9%2A5.67%2A10%5E%7B-8%7D%28T_p_1%5E4-308%5E4%29%5D-10%28T_p_1-308%29%20-%20%280.553-0.001T_p_1%290.83%2A700%20%3D%200%5C%5C%5C%5C3798.94-5.103%2A10%5E%7B-8%7DT_p_1%5E4%20-%209.419T_p_1%20%3D%200%5C%5C%5C%5CApply%20%5C%20%20%5C%20iteration%20%5C%20method%20%5C%20to%20%5C%20solve%20%5C%20for%20%5C%20T_p_1%5C%5C%5C%5CT_p_1%20%3D%20335.05%20%5C%20k)

(b)the electrical power generated for a breezy winter day

![[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k](https://tex.z-dn.net/?f=%5B0.83%2A700-0.9%2A5.67%2A10%5E%7B-8%7D%28T_p_2%5E4-258%5E4%29%5D-10%28T_p_2-258%29%20-%20%280.553-0.001T_p_2%290.83%2A700%20%3D%200%5C%5C%5C%5C8225.81-5.103%2A10%5E%7B-8%7DT_p_2%5E4%20-%2029.419T_p_2%20%3D%200%5C%5C%5C%5CApply%20%5C%20%20%5C%20iteration%20%5C%20method%20%5C%20to%20%5C%20solve%20%5C%20for%20%5C%20T_p_2%5C%5C%5C%5CT_p_2%20%3D%20279.6%20%5C%20k)

I think you're saying that once you start pushing on the cars, you want to be able to stop each one in the same time.
This is sneaky. At first, I thought it must be both 'c' and 'd'. But it's not
kinetic energy, for reasons I'm not ambitious enough to go into.
(And besides, there's no great honor awarded around here for explaining
why any given choice is NOT the answer.)
The answer is momentum.
Momentum is (mass x speed). Change in momentum is (force x time).
No matter the weight (mass) or speed of the car, the one with the greater
momentum is always the one that will require the greater (force x time)
to stop it. If the time is the same for any car, then more momentum
will always require more force.