Answer:
Explanation:
change in momentum = m ( v - u ) ; m is mass , V is final velocity and u is initial velocity .
= 250 ( 28 - 6 )
= 5500. N.s
for calculating magnitude of force we shall apply the concept of impulse
Impulse = F X t = m( v-u ) ; F is force applied for time t .
F x 60 = 5500
F = 91.67 N .
force exerted = 91.67 N .
Answer:
a = αR
Then We apply the Newton's second law of motion:
∑ F = ma
F - mg sinθ - f = ma.
F - mg sinθ - μmg cosθ = ma
Using given data we have:
F - 3217 = 470a ........................... (1)
Apply the Newton's law for rotation:
∑τ = I α
FR - (mg sinθ)R = 
then: F - mg sinθ = 3ma / 2.
F - 2774 = 705a .......................... (2)
solving 1 and 2 we get:
F = 4103 N
a = 1.885 m/s²
b) In this case the time is given by:
t =
it comes from: d =
starting from rest vo = 0
t =
= 3.09 s
We should first calculate the highest point that ball reaches.
y' = 40 - 32t = 0
t = 40/32 = 1,25s
y = 25 feet.
To calculate average velocity we use simple formula:
Vav=s/t where s is traveled distance by the time t.
for t=2 we calculate y
y = 16
(i) for t = 2,5 y = 0
Vav = 16/0.5 = 32 feet/s
(ii) for t = 2.1 y = 13.44
Vav = 2.56/0.1 = 25.6 feet/s
(iii) for t = 2.01 y = 15.7584
Vav = 0.2416/0.01= 24.16 feet/s
it seems like the answer would be 24 feet/s. There is a way to calculate that.
Answer
given,
mass of bowling ball = 7.25 Kg
moving speed of the bowling ball = 9.85 m/s
mass of bowling in = 0.875 Kg
scattered at an angle = θ = 21.5°
speed after the collision = 10.5 m/s
angle of the bowling ball
![tan \theta_1 = \dfrac{-[m_2v_2Sin \theta_2]}{m_1v_1 - (m_2v_2cos \theta_2)}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5Bm_2v_2Sin%20%5Ctheta_2%5D%7D%7Bm_1v_1%20-%20%28m_2v_2cos%20%5Ctheta_2%29%7D)
![tan \theta_1 = \dfrac{-[0.875\times 10.5 \times Sin 21.5^0]}{7.25\times 9.85 - (0.875\times 10.5 \times cos 21.5^0)}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5B0.875%5Ctimes%2010.5%20%5Ctimes%20Sin%2021.5%5E0%5D%7D%7B7.25%5Ctimes%209.85%20-%20%280.875%5Ctimes%2010.5%20%5Ctimes%20cos%2021.5%5E0%29%7D)
![tan \theta_1 = \dfrac{-[3.3672]}{62.86}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5B3.3672%5D%7D%7B62.86%7D)


b) magnitude of final velocity


v = 8.68 m/s
<span>1 cal = 4,185 J
1 kcal = 1*10^3 cal
or
=1000 cal</span>