<h2>It will take 0.125 seconds to reach the net.</h2>
Explanation:
Initial speed, u = 34 ft/s = 10.36 m/s
Acceleration, a = -9.81 m/s²
Displacement, s = Final height - Initial height = 8 - 4 = 4 ft = 1.22 m
We have equation of motion, s = ut + 0.5 at²
Substituting
s = ut + 0.5 at²
1.22 = 10.36 x t + 0.5 x -9.81 x t²
4.905t² - 10.36 t + 1.22 = 0
t = 1.99 s or t = 0.125 seconds
Minimum time is 0.125 seconds.
It will take 0.125 seconds to reach the net.
Which statement is always false for athletes participating in team sports?
Answer: Out of all the options shown above the one that best represents the statement that is alway false for athletes participating in team sports is answer choice C) Conflict resolution is a sign of poor sportsmanship. All the other choices are true when it comes to team sports.
I hope it helps, Regards.
Answer:
statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.
Explanation:
The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.
The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction
here, the downward direction signifies the downward motion parallel to the inclined plane.
Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.
Hence, for the block to stop sliding the the above statement should be true.
Part a)
per day electricity power consumed when 100 W bulb is used for 8 hours
for one year consumption
now the cost will be given
now when other energy efficient light is used
for one year consumption
now the cost will be given
Answer:
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.
how much work is done on the monitor by (a) friction, (b) gravity
work(friction) = 453.5J
work(gravity) = -453.5J
Explanation:
Given that,
mass = 14kg
displacement length = 5.50m
displacement angle = 36.9°
velocity = 2.30cm/s
F = ma
work(friction) = mgsinθ .displacement
= (14) (9.81) (5.5sin36.9°)
= 453.5J
work(gravity)
= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)
= 126.9°
work(gravity) = (14) (9.81) (5.5cos126.9°)
= -453.5J
