Answer:
Thus, the payback time for cavity wall insulation is 7.5 years
Actual Mechanical Advantage(AMA) = Weight / Force
Here, Weight = 764 N
Force = 255 N
Substitute the values in to the expression,
AMA = 764 / 255
AMA = 2.99
After rounding-off to the nearest tenth value, it would be 3
Finally, option C would be your answer.
Hope this helps!
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
Answer:
The energy that the truck lose to air resistance per hour is 87.47MJ
Explanation:
To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by
Our values are:
Replacing,
We need calculate now the energy lost through a time T, then,
But we know that d is equal to
Where
v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,
(per hour)
Therefore the energy that the truck lose to air resistance per hour is 87.47MJ
Answer:
7.24 Ω
Explanation:
Power = energy / time = 195 J / 9.81 = 19.88 W
and power = V² / R
R resistance = 12 V² / 19.88 W = 7.24 Ω
