Answer:
Use the method on the image and solve it.
Answer:
1. 3 m
2. 27 s
Explanation:
1. "A car traveling at +33 m/s sees a red light and has to stop. If the driver can accelerate at -5.5 m/s², how far does it travel?"
Given:
v₀ = 33 m/s
v = 0 m/s
a = -5.5 m/s²
Unknown: Δx
To determine the equation you need, look for which variable you don't have and aren't solving for. In this case, we aren't given time and aren't solving for time. So look for an equation that doesn't have t in it.
Equation: v² = v₀² + 2aΔx
Substitute and solve:
(0 m/s)² = (33 m/s)² + 2(-5.5 m/s²) Δx
Δx = 3 m
2. "A plane starting from rest at one end of a runway accelerates at 4.8 m/s² for 1800 m. How long did it take to accelerate?"
Given:
v₀ = 0 m/s
a = 4.8 m/s²
Δx = 1800 m
Unknown: t
Equation: Δx = v₀ t + ½ a t²
Substitute and solve:
1800 m = (0 m/s) t + ½ (4.8 m/s²) t²
t ≈ 27 s
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)

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.
Now from Newton`s first law, the net force on gymnast,

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast (
acceleration due to gravity)
Therefore,
OR 
Given
and
Substituting these values in above formula and calculate the force exerted by the gymnast,

