Answer:
F = 800 [N]
Explanation:
To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.
We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.
ΣPbefore = ΣPafter
where:
m₁ = mass of the hammer = 0.15 [m/s]
v₁ = velocity of the hammer = 8 [m/s]
F = force [N] (units of Newtons)
t = time = 0.0015 [s]
v₂ = velocity of the hammer after the impact = 0
![(0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N]](https://tex.z-dn.net/?f=%280.15%2A8%29-%28F%2A0.0015%29%20%3D%20%280.15%2A0%29%5C%5CF%2A0.0015%20%3D%200.15%2A8%5C%5CF%20%3D%201.2%2F%280.0015%29%5C%5CF%20%3D%20800%20%5BN%5D)
Note: The force is taken as negative since it is exerted by the nail on the hammer and this force is directed in the opposite direction to the movement of the hammer.
Answer:
U_f = (U_o)/2)
Explanation:
The capacitance of a given capacitor is given by the formula;
C = ε_o•A/d
While energy stored in plates capacitor is given as; U_o = Q²/2C
Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;
C' = ε_o•4A/2d
Hence, C' = 2C
so capacitance is now doubled
Thus, the final energy stored between the plates of capacitor is given as;
U_f = Q²/2C'
From earlier, we saw that C' = 2C.
Thus;
U_f = Q²/2(2C)
U_f = Q²/4C
Rearranging, we have;
U_f = (1/2)(Q²/2C)
From earlier, U_o = Q²/2C
Hence,
U_f = (1/2)(U_o)
Or
U_f = (U_o/2)
Answer:
3.025 revolutions
Explanation:
To solve this question firstly we need to calculate the time t
Given the angular velocity and the angular acceleration
The time t = angular velocity/angular acceleration
Converting 50rpm to rad/s
= 50 × 2π /60
=1.67 rad/s
t = 1.67/0.46
t =3.63s
The wheel makes 50 revolutions in one minute
Therefore in 3.63s, it will make
3.63 × 50/60
= 181.5/60
=3.025 revolutions
Approximately 3 revolutions
