<h2>Book will hit the foot with force 21.8 N</h2>
Explanation:
The momentum of any object = m v
here m is the mass of body and v is its velocity
Thus momentum P = 3.2 x 1 = 3.2 kg-m
As the book is falling freely , so it will fall with and acceleration g
The force exerted by it is = m g
Thus force = 2.18 x 10 = 21.8 N
I would say friction because it requires to surfaces in order for the force to take place
correct me if I'm wrong.
Answer:
The ability to react to a certain stimulus with a speedy and effective manner
Explanation:
Answer:
64.945 miles per hour
Explanation:
Since the frequency of sound heard is higher than actual frequency, the ambulance is moving towards you!
The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by
+f = f₀/[1 - (v₀/v)]
+f = frequency of sound as heard from the distance away = 8.61 KHz
f₀ = real frequency of sound = 7.87 KHz
v₀ = velocity at which the sound source is moving towards the reference point = ?
v = velocity of sound waves = 343 m/s
8.61 = 7.87/(1 - (v₀/v))
1 - (v₀/343) = 0.9141
v₀/343 = 1 - 0.9141 = 0.0859
v₀ = 343 × 0.0859 = 29.48 m/s = 64.945 miles per hour
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
