A farm tractor tows a 3300-kg trailer up a 14" incline with a steady speed of 2.8 m/s. what force does the tractor exert on the
1 answer:
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The centripetal acceleration a is 4.32 10^-4 m/s^2.
<u>Explanation:</u>
The speed is constant and computing the speed from the distance and time for one full lap.
Given, distance = 400 mm = 0.4 m, Time = 100 s.
Computing the v = 0.4 m / 100 s
v = 4 10^-3 m/s.
radius of the circular end r = 37 mm = 0.037 m.
centripetal acceleration a = v^2 / r
= (4 10^-3)^2 / 0.037
a = 4.32 10^-4 m/s^2.
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Explanation :
It is given that,
In the given figure all the three resistors are in series.
Current flowing in the circuit,
Voltage,
We know that in series circuit the current flowing in all resistors is same.
Using Ohm's law, we get:
Hence, this is the required solution.