210J
PE is mgh in this context.
i don't know the anss , sorry.
Answer:
g'(10) = 
Explanation:
Since g is the inverse of f ,
We can write
g(f(x)) = x <em> </em><em>(Identity)</em>
Differentiating both sides of the equation we get,
g'(f(x)).f'(x) = 1
g'(10) = 
--equation[1] Where f(x) = 10
Now, we have to find x when f(x) = 10
Thus 10 = 
+ 2
= 8
x = 
Since f(x) = + 2
f'(x) = -
f'() = -4 × 4 = -16
Putting it in equation 1, we get:
We get g'(10) = -
Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²
Answer:
-35 m/s
Explanation:
Momentum is conserved.
Momentum before firing = momentum after firing
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Before the bullet is fired, the bullet and rifle have no velocity, so u₁ and u₂ are 0.
0 = m₁v₁ + m₂v₂
Given m₁ = 0.7 kg, v₁ = 350 m/s, and m₂ = 7 kg:
0 = (0.7 kg) (350 m/s) + (7 kg) v
v = -35 m/s
The rifle recoils at 35 m/s in the opposite direction.
