Answer: 12.5 km/s
I don't really know how to explain this, but here is your answer.
Answer:C
Explanation:
When a constant horizontal force is applied to the box, box started moving in the horizontal direction such that it moves with constant velocity 
Constant velocity implies that net force on the box is zero
i.e. there must be an opposing force which is equal to the applied force and friction force can serve that purpose.
So option c is the correct choice.
Answer: 90.1 s
Explanation:
Use equation for power:
P=F*V
Use eqation for force:
F=ma
F---force
V---velocity
Vr=om/s
V=30m/s
m=1000kg
P=10000W
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P=FV
F=P/V
F=10000W/30m/s
F=333.33N
Use equation for force to find accelartaion.
F=ma
a=F/m
a=333.33N/1000kg
a=0.333 m/s²
Use equation for accelaration to find out time:
a=(V-Vs)/t
t=(V-Vs)/a
t=(30m/s)/(0.333m/s²)
t=90.09 s≈90.1 s
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|acceleration| = (change in speed) / (time for the change)
Change in the car's speed = (27 - 0) = 27 m/s
Time for the change = 10 sec
|acceleration| = (27 m/s) / (10 s) = 2.7 m/s² .
That's the magnitude of the car's acceleration.
We don't know anything about its direction.
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2