Answer:
Explanation:
Remark
In general, these 3rd class levers are very inefficient. Because the force distance is smaller than the load distance, you need to pull upward with more force that the weight of the load. So whatever the load is, the force is going to be much greater.
The distances are always measured to the pivot unless you are asked something specific otherwise.
Givens
F = ?
weight = 6N
Force Distance = F*d = 0.5 m
Weight Distance =W*d1 = 2 m
Formula
F*Fd = W*Wd
Solution
F*0.5 = 6 * 2 Divide by 0.5
F = 12/0.5
F = 24 N upwards
Solving this using the time, we know that range = horizontal velocity x time of flight
since
there are no horizontal forces acting on the ball, there are no
horizontal accelerations and the initial horizontal velocity of 36 cos
28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have
range = 36 cos 28 x 3.44 s = 109.3 m
<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3>
<h3>B )
The force applied on the smaller gear is transmitted without any loss to the larger gear .</h3><h3 /><h3>
C ) the direction of motion can be changed without changing the direction of the applied force .</h3>
D ) the system would continue to move without any further, after and initial force has set in motion.
Old temperature = 283 K.
New temp = 323 K.
(323/283) x (325 kPa) = 371 kPa.
Answer:
1700 Joules
Explanation:
Work=force x distance
Force = 170 kg
Distance= 10 Meters
170 x 10 = 1700 Joules of work
