Take 68.2/60 = 1.137 hr
take 56.9/1.137 = 50.043 mi/hr
take 189/211 = 0.896
24.8/2 = 12.4 m
12.4/82.3 = 0.15s
Answer:
0.2687 approximately 0.27
Explanation:
Diameter = 0.320
Speed = 40.0 rev/min
We are required to find coefficient of static friction between friction and button
The radius can be calculated as
0.320/2
= 0.160m
Then we have the rotational speed w = 40rev/min x 2pi/60
= 4.19 rad/s
umg = mrw²
u = mrw²/mg
u = rw²/g -------(1)
g = 9.8
When we put values into equation 1
0.150m x 4.19² / 9.8
= 0.150m x 17.5561 /9.8
= 0.2689
This is approximately 0.27
Answer:
C) 3,000 kg m/s
Explanation:
We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.
The vertical velocity of the motorcycle at time t is given by (free-fall motion):

where
is the initial vertical velocity (zero, since the motorcycle is not moving)
g = 9.8 m/s^2 is the acceleration due to gravity
t is the time
Since the motorcycle hits the ground after t = 3 seconds, we have

And since we know its mass, m=100 kg, we can find its momentum:

and the negative sign simply means downward direction.
Answer:0.253Joules
Explanation:
First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.
F = ke where;
F is the force
k is spring constant = 34N/m
e is the extension = 0.12m
F = 34× 0.12 = 4.08N
To get work done,
Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.
Work done = Force × Distance
Since F = 4.08m, distance = 0.062m
Work done = 4.08 × 0.062
Work done = 0.253Joules
Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules
Answer:
hello your question is not properly arranged attached below is the arranged table and solution
answer : attached table below
Explanation:
Given data:
02 molecules size = 10^-10m
smoke particles size = 0.3 mm
cloud droplets size = 20 mm
Rain droplets size = 3 mm
Attached below is a table showing the kind of scattering that is expected to occur at various wave lengths
Note : For Rayleigh scattering the wave particle is smaller than the wave length while for Non-selective scattering the wave particle is greater than the wavelength.
and For Mie scattering the wavelength is the same as the wavelength.