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
ball drops 45m under g=10m/s/s
45=1/2x10xt^2 ... application of kinematic equaion from rest
90/10=t^2
t=3
24.0 m in 3 secs => 8m/s no air resistance
Hello!
Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:
← The circumference of the orbit
speed = orbital speed, we will solve for this later
time = period
Therefore:
Where 'r' is the orbital radius of the satellite.
First, let's solve for 'v' assuming a uniform orbit using the equation:
G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)
m = mass of the earth (5.98 × 10²⁴ kg)
r = radius of orbit (1.276 × 10⁷ m)
Plug in the givens:
Now, we can solve for the period:
Answer:
True
Explanation: If this is a true or false question it is *T*
