Answer:
Explanation:
When two people carry a weight placing it on a very light board then we neglect the weight of the board and the board acts like a beam with two mass placed over it and two end point supports are here provided by men.
Then balancing the forces in the vertical direction, weight of the motor is:
These men are able to carry the motor in equilibrium condition because they have placed the motor (considering it a point mass) at a position such that the moment due to the lifting forces are equal and opposite.
Answer" The average force on the wall be equal to 0.5 N.
Explanation:
To calculate the average force on the wall, we use the following equation:
where,
F = average force = ?N
m = mass of the ball = 0.05kg
= Initial velocity of the ball = -10m/s (negative sign because the ball rebounds)
= Final velocity of the ball = 10m/s
t = time taken by the ball = 2s
Putting values in above equation, we get:
![F=\frac{0.05kg[10-(-10)]m/s}{2s}\\\\F=\frac{0.05kg(20m/s)}{2s}\\\\F=0.5kgm/s^2=0.5N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B0.05kg%5B10-%28-10%29%5Dm%2Fs%7D%7B2s%7D%5C%5C%5C%5CF%3D%5Cfrac%7B0.05kg%2820m%2Fs%29%7D%7B2s%7D%5C%5C%5C%5CF%3D0.5kgm%2Fs%5E2%3D0.5N)
Hence, the average force on the wall will be equal to 0.5N.
I would say C! But I’m not 100% sure.
The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.
<h3>Work done by the applied force</h3>
The area under force versus displacement graph is work done.
The total work done by pulling the object 7 m, can be grouped into two areas;
- First area, A1 = area of triangle from 0 m to 2.0 m
- Second area, A2 = area of trapezium, from 2.0 m to 7.0 m
A1 = ¹/₂ bh
A1 = ¹/₂ x (2) x (20)
A1 = 20 J
A2 = ¹/₂(large base + small base) x height
A2 = ¹/₂[(7 - 2) + (7-3)] x 50
A2 = ¹/₂(5 + 4) x 50
A2 = 225 J
<h3>Total work done </h3>
W = A1 + A2
W = 20 J + 225 J
W = 245 J
Learn more about work done here: brainly.com/question/8119756
Answer:
a) 20s
b) 500m
Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.
To find time, we apply the UARM formula:
v final = (a x t) + v initial
Replacing the values gives us:
0 = (-10 x t) + 100
-100 = -10t
t = 10s
It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur
So 10s going up and another 10s going down:
10x2 = 20s
b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:
Δy = (1/2)(a)(t^2) + (v initial)(t)
Replacing the values gives us:
Δy = (1/2)(-10)(10^2) + (100)(10)
= (-5)(100) + 1000
= -500 + 1000
= 500 m
Hope this helps, brainliest would be appreciated :)
