<span>vf^2 = vi^2 + 2*a*d
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vf = velocity final
vi = velocity initial
a = acceleration
d = distance
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since the airplane is decelerating to zero, vf = 0
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0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters
</span>
<span>The change in internal energy is only gravitional PE because the tube is being drug up at a constant speed. Since it is at a constant speed, the change in KE is 0.
Change in PE = m*g*h = 78 kg * 10 m/s^2 * 30 m = 23400 J
Work done on the system is from the force
Work = force * distance = 350 N * 120 m = 42000 J
So, work added 42000 J to the system, but the rider's energy only increased 23400 J. Therefore, friction took up the difference. Friction is where the thermal energy comes from
Q = 42000 J - 23400 J = 18600 J.
Therfore, friction generated 18600 J of heat to the surroundings.</span>
Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )
Given:
(Initial velocity)u=20 m/s
At the maximum height the final velocity of the ball is 0.
Also since it is a free falling object the acceleration acting on the ball is due to gravity g.
Thus a=- 9.8 m/s^2
Now consider the equation
v^2-u^2= 2as
Where v is the final velocity which is measured in m/s
Where u is the initial velocity which is measured in m/s
a is the acceleration due to gravity measured in m/s^2
s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.
Substituting the given values in the above formula we get
0-(20x20)= 2 x- 9.8 x s
s= 400/19.6= 20.41m
Thus the maximum height attained is 20.41 m by the ball
Answer:
The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).
Explanation:
Hi there!
The equation for the velocity of an object thrown upward is the following:
v = v0 + g · t
Where:
v = velocity of the ball.
v0 = initial velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
t = time.
To find the velocity of the ball at t = 0.40 s, we have to replace "t" by 0.40 s in the equation:
v = v0 + g · t
v = 3.0 m/s - 9.8 m/s² · 0.40 s
v = -0.92 m/s
The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).
