Answer:
W = 172.5 J
Explanation:
given,
mass of the fruit crate = 14.5 kg
initial velocity to lift = 0.500 m/s
increase in the tension = 150 N
lift of crate = 1.15 m
work done by the tension = ?
work done = force x displacement
W = F s cos θ
θ = 0°
W = F s x cos 0
W = 150 x 1.15 x 1
W = 172.5 J
Work done on the crate by the tension force = W = 172.5 J
Let Karen's forward speed be considered as positive.
Therefore, before the headband is tossed backward, the speed of the headband is
V = 9 m/s
The headband is tossed backward relative to Karen at a speed of 20 m/s. Therefore the speed of the headband relative to Karen is
U = -20 m/s
The absolute speed of the headband, relative to a stationary observer is
V - U
= 9 + (-20)
= - 11 m/s
Answer:
The stationary observes the headband traveling (in the opposite direction to Karen) at a speed of 11 m/s backward.
By law of refraction we know that image position and object positions are related to each other by following relation
here we know that
now by above formula
so apparent depth of the bottom is seen by the observer as h = 3.39 cm
Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
