I think is near the ozone layer, i am not sure
Answer:
The rate of change of the shadow length of a person is 2.692 ft/s
Solution:
As per the question:
Height of a person, H = 20 ft
Height of a person, h = 7 ft
Rate = 5 ft/s
Now,
From Fig.1:
b = person's distance from the lamp post
a = shadow length
Also, from the similarity of the triangles, we can write:
Differentiating the above eqn w.r.t t:
Now, we know that:
Rate = 
Thus
Answer:
<h2> $1.50</h2>
Explanation:
Given data
power P= 2 kW
time t= 15 min to hours = 15/60= 1/4 h
cost of power consumption per kWh= 10 cent = $0.1
We are expected to compute the cost of operating the heater for 30 days
but let us computer the energy consumption for one day
Energy of heater for one day= 2* 1/4 = 0.5 kWh
the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50
<u><em>Hence it will cost $1.50 for 30 days operation</em></u>
Answer:
a) 35.75 ft/s
b) 45 ft
Explanation:
<u>Given </u>
Weight W = 100 lbf
mass(m) = 100*32.174/32.2=99.92 lb
decrease in kinetic energy ΔKE = -500 ft.lbf
increase in kinetic energy ΔPE= 1500 ft.lbf
initial velocity V_1 = 40 ft/s
initial height h_1 = 30 ft/s
The gravitational acceleration g = 32.2 ft/s2 Required
(a) Final velocity V_2 (a) Final elevation h_2
<u>Solution </u>
Change in kinetic energy is defined by
ΔKE = .5*m *( V_2 ^2-V_1^2)
Change in potential energy is defined by
ΔKE = W *( h_2 -h_1 )
Then,
-500=.5*99.92*1/32.174*(V_2 ^2-40^2)
V_2=35.75 ft/s
1500 = 100 x (h_2 — 30)
h_2= 45 ft
Answer:
If you add the 2 velocity vectors you get the velocity of the third vector which is the (change in) velocity of the cue ball after striking the cushion
V * 2 * cos 45 = 4 * cos 45 = 2.83 m/s
