Answer:
D) Higher Frequency
Explanation:
Higher frequency because it says a vehicle “towards” the officer
6.
A. 1575 - 1265 = 310J
B. KE 1/2 MV^2
V=√2·KE/M = √2(310)/12 V = 7.2mls
C. PE = 1265 = mgh
h= 1265/mg = 1265/(12)(92) h= 10.8m
7.
A. KE = 1/2 mv^2 0.5(5)(12)^2 KE = 360J
B. PE = mgh = (5)(9.8)(2.6) PE = 127.4J
C. ME = KE + PE = 360 + 127.4 ME = 487.4J
Apply conservation of angular momentum:
L = Iw = const.
L = angular momentum, I = moment of inertia, w = angular velocity, L must stay constant.
L must stay the same before and after the professor brings the dumbbells closer to himself.
His initial angular velocity is 2π radians divided by 2.0 seconds, or π rad/s. His initial moment of inertia is 3.0kg•m^2
His final moment of inertia is 2.2kg•m^2.
Calculate the initial angular velocity:
L = 3.0π
Final angular velocity:
L = 2.2w
Set the initial and final angular momentum equal to each other and solve for the final angular velocity w:
3.0π = 2.2w
w = 1.4π rad/s
The rotational energy is given by:
KE = 0.5Iw^2
Initial rotational energy:
KE = 0.5(3.0)(π)^2 = 14.8J
Final rotational energy:
KE = 0.5(2.2)(1.4)^2 = 21.3J
There is an increase in rotational energy. Where did this energy come from? It came from changing the moment of inertia. The professor had to exert a radially inward force to pull in the dumbbells, doing work that increases his rotational energy.
Answer:
Time of flight = 4.08seconds
Horizontal component of initial velocity is 17.32m/s
Explanation: complete question( and the horizontal component of the initial velocity.)
The equation for time of flight of a projectile is given as T= 2u/g
T=( 2×20)/9.8
T= 40/9.8= 4.08seconds
Horizontal component of initial velocity Vix= Vi Costheta
Vix= 20× cos 30°
Vix= 20×0.8660
Vix= 17.32m/s
At standard atmospheric pressure, 100° C
is defined as the boiling point of water.