It will take at least 3 hours for them to get to Bridgeport
Answer:
The mass of the ball is 0.23 kg
Explanation:
Given that
radius ,r= 3.74 cm
Density of the milk ,ρ = 1.04 g/cm³ = 1.04 x 10⁻³ kg/cm³
Normal force ,N= 9.03 x 10⁻² N
The volume of the ball V
V= 219.13 cm³
The bouncy force on the ball = Fb
Fb = ρ V g
Fb + N = m g
m=Mass of the ball = Density x volume
m = γ V , γ =Density of the Ball
ρ V g + N = γ V g ( take g= 10 m/s²)
γ = 0.00108 kg/cm³
m = γ V
m = 0.00108 x 219.13
m= 0.23 kg
The mass of the ball is 0.23 kg
Explanation:
Below is an attachment containing the solution.
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
The answer would be point A.
Hope this helped you.
