Answer:
t = 2 s
Explanation:
In order to find the time taken by the stone to fall from the top of the building to the ground we can use 2nd equation of motion. 2nd equation of motion is as follows:
s = Vit + (0.5)gt²
where,
t = time = ?
Vi = Initial Velocity = 20 m/s
s = height of building = 60 m
g = 9.8 m/s²
Therefore,
60 m = (20 m/s)t + (0.5)(9.8 m/s²)t²
4.9t² + 20t - 60 = 0
solving this quadratic equation we get:
t = -6.1 s (OR) t = 2 s
Since, the time cannot be negative in magnitude.
Therefore,
<u>t = 2 s</u>
In the conservation of mass, mass is never created or destroyed in chemical reactions in the same way water is not created or destroyed it is only transferred from one form to another and its mass is always conserved.
Answer:
Part a)
![x = 0.4 m](https://tex.z-dn.net/?f=x%20%3D%200.4%20m)
Part b)
![v_i = 11.7 m/s](https://tex.z-dn.net/?f=v_i%20%3D%2011.7%20m%2Fs)
Part c)
Speed is more than the required speed so it will reach the top
Explanation:
Part a)
As we know that there is no frictional force while block is moving on horizontal plane
so we can use energy conservation on the block
![\frac{1}{2}mv^2 = \frac{1}{2}kx^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
![\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D0.500%2812%5E2%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28450%29x%5E2)
![x = 0.4 m](https://tex.z-dn.net/?f=x%20%3D%200.4%20m)
Part b)
If the track has average frictional force of 7 N then work done by friction while block slides up is given as
![W_f = -7( \pi R)](https://tex.z-dn.net/?f=W_f%20%3D%20-7%28%20%5Cpi%20R%29)
![W_f = -7(\pi \times 1.00)](https://tex.z-dn.net/?f=W_f%20%3D%20-7%28%5Cpi%20%5Ctimes%201.00%29)
![W_f = -22 J](https://tex.z-dn.net/?f=W_f%20%3D%20-22%20J)
work done against gravity is given as
![W_g = - mg(2R)](https://tex.z-dn.net/?f=W_g%20%3D%20-%20mg%282R%29)
![W_g = -(0.500)(9.8)(2\times 1)](https://tex.z-dn.net/?f=W_g%20%3D%20-%280.500%29%289.8%29%282%5Ctimes%201%29)
![W_g = -9.8 J](https://tex.z-dn.net/?f=W_g%20%3D%20-9.8%20J)
Now by work energy equation we have
![\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv_i%5E2%20%2B%20W_f%20%2B%20W_g%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv_f%5E2)
![\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D0.5%2812%5E2%29%20-%209.8%20-%2022%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.5%29v_f%5E2)
![v_f = 4.1 m/s](https://tex.z-dn.net/?f=v_f%20%3D%204.1%20m%2Fs)
Part c)
now minimum speed required at the top is such that the normal force must be zero
![mg = \frac{mv^2}{R}](https://tex.z-dn.net/?f=mg%20%3D%20%5Cfrac%7Bmv%5E2%7D%7BR%7D)
![v = \sqrt{Rg}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7BRg%7D)
![v = 3.13 m/s](https://tex.z-dn.net/?f=v%20%3D%203.13%20m%2Fs)
so here we got speed more than the required speed so it will reach the top
Are you answering a question or asking? You have already seemed to get the answer, A is the correct answer :I
Answer:
The ball has no momentum
Explanation:
The given parameters are;
The mass of the ball = 5 kg
The velocity of the ball = 0 (The ball is sitting on the floor without moving)
The momentum of the ball = The mass of the ball × Velocity of the ball
Therefore, the momentum of the ball = 5 kg × 0 m/s = 0
The momentum of the ball is zero, the ball has no momentum.