A. Units
the rest can be understand that he did not clarify
Answer:
a) -2.1731 m⁻s
Explanation:
L= length of thread, H= height of ball from ceiling,
N/B h= l cosФ, Ф=31°, l=2.7m, g=9.8m⁻s, m= 6kg
Required to find speed v, of the ball in m⁻s
ω=√g/h, =√g/lcosФ
⇒ ω=√9.8/2.7×cos31°
= 1.992rad⁻s
Now, speed in m/s, v= r.ω
where tanФ= r/h ⇒
r= h×cos 31°
r= 2.7 × cos(31) × tan(31) =-1.0909m
v= 1.992× -1.0909 = -2.1731 m/s
Answer: The recoil speed is - 8.9604. m/s.
Explanation: According to the Third Law of Newton, every action has an oppsite and equal reaction, and the Second Law of Newton, Force=mass·acceleration. Acceleration is a variation in velocity by any given time, so Force = mass·velocity·time.
Combining the two laws, there is : m1·v1 = - m2·v2. This is the Law of Conservation of Momentum.
Substituting and calculating:
v2 = - () · v1
v2 = - · 5.24.
v2 = - 8.9604.
The recoil speed of the thorium nucleus is - 8.9604.m/s.
Answer:
79.98
if this is a simple math problem, the (•) signifies multiplication
Kenetic energy = (1/2)(mass)(velocity)^2
Simply plug in values.
(0.5)(10kg)(2m/s) = 10 Joules; A