Answer:
Explanation:
moment of inertia of each blade which is similar to rod rotating about its one end
= 1/3 ml²
moment of inertia of 3 blades = ml²
= 5500 x 46²
I = 11638 x 10³ kg m²
angular velocity = 2πn where n is rotation per second
n = 11 / 60
angular velocity = 2π x 11/60
= 1.1513 rad /s
angular momentum
= moment of inertia x angular velocity
= 11638 x 10³ x 1.1513
= 13399 x 10³ kg m² per second.
Answer:
9.47 rad/s^2
Explanation:
Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,
s = 35.4 m
let a be the linear acceleration.
Use III equation of motion.
v^2 = u^2 + 2 a s
7.1 x 7.1 = 0 + 2 x a x 35.4
a = 0.71 m/s^2
Now the relation between linear acceleration and angular acceleration is
a = r x α
where, α is angular acceleration
α = 0.71 / 0.075 = 9.47 rad/s^2
<h2>
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>
Explanation:
One period means time taken to complete one revolution.
In case of swings in one period time it travels the same position through two times.
Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.
For 1 period = 2 times
For 3 periods = 3 x For 1 period
For 3 periods = 3 x 2 times
For 3 periods = 6 times
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.
Answer:
Option 5. 1 and 3
Solution:
The only forces acting on the tennis ball after it has left contact with the racquet and the instant before it touches the ground are the force of gravity in the downward direction and the force by the air exerted on the ball.
The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.
In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction before it hits the ground.
Answer:
Difference in height = 7.5 cm
Explanation:
We are given;.
Height of ethyl alcohol;h2 = 20 cm = 0.2 m
Density of glycerin: ρ1 = 1260 kg/m³
Density of ethyl alcohol; ρ2 = 790 kg/m³
To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.
Thus;
P1 = P2
Formula for pressure is; P = ρgh
Thus;
ρ1 × g × h1 = ρ2 × g × h2
g will cancel out to give;
ρ1 × h1 = ρ2× h2
Making h1 the subject, we have;
h1 = (ρ2× h2)/ρ1
h1 = (790 × 0.2)/1260
h1 = 0.125 m
Difference in height will be;
Δh = h2 - h1
Δh = 0.2 - 0.125
Δh = 0.075 m = 7.5 cm
