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
Answer:
The correct answer to the following question will be Option A (moment arm; pivot point).
Explanation:
- The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
- The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
- The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.
The other three choices are not related to the given situation. So that option A is the appropriate choice.
Answer:
Explanation:
Given
Force P is acting upward
C is vertical contact Force
W is the weight of the crate
As P is unable to move the Block therefore Normal reaction keeps on acting on block
thus we can say that
P-W+C=0
P=W-C
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
Answer:
The centripetal force acting on the skater is <u>48.32 N.</u>
Explanation:
Given:
Radius of circular track is, 
Tangential speed of the skater is, 
Mass of the skater is, 
We are asked to find the centripetal force acting on the skater.
We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.
Centripetal force acting on the skater is given as:
Now, plug in the given values of the known quantities and solve for centripetal force, 
. This gives,
Therefore, the centripetal force acting on the skater is 48.32 N.
