For this case, the first thing we must do is define a reference system.
Suppose that the positive direction of the reference system is upward.
We have that the sum of forces in the vertical axis is given by:
Fy = Fp - Fg
Substituting values:
Fy = 5500 - 6000
Fy = - 500
The negative sign means that the direction of the force with respect to the defined coordinate system is downward.
Answer:
The net force is:
↓ 500N
Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
Work=applied Force x distance
= 1275 x 26
=33150 Joules
Answer:
%
Explanation:
We have given cooler temperature that is 
= 6°C =273+6=279 K
And the warm temperature of water that is 
The maximum possible efficiency is given by 
So maximum efficiency 
%
Acceleration = velocity/ time
Acceleration = 0.7-0.3 /30= 0.01 m/s^2
Notice that velocity is calculated the final speed minus the initial !
