Answer:
0.36
Explanation:
The maximum force of friction exerted by the surface is given by:
(1)
where
is the coefficient of friction
N is the normal reaction
The shed's weight is 2200 N. Since there is no motion along the vertical direction, the normal reaction is equal and opposite to the weight, so
N = 2200 N
The horizontal force that is pushing the shed is
F = 800 N
In order for it to keep moving, the force of friction (which acts horizontally in the opposite direction) must be not greater than this value. So the maximum force of friction must be
And substituting the values into eq.(1), we can find the maximum value of the coefficient of friction:
Answer:
Distance, d = 3242.19 meters
Explanation:
It is given that,
Speed of the dog, v = 19.85 mi/hr
Since, 1 mph = 0.44704 m/s
v = 19.85 mi/hr = 8.873 m/s
Time taken by the dog, t = 6.09 min = 365.4 sec
Let the distance covered by the dog during this time period is d. It can be calculated by the speed of the dog multiplied by the time taken as :
d = 3242.19 meters
So, the the dog travel during this time period is 3242.19 meters. Hence, this is the required solution.
Answer: hello your question is incomplete below is the missing part
A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.
answer:
To 2 significant Figures = 1.6 m/s^2
Explanation:
<u>Calculate the magnitude of minimum acceleration at which the thief can descend </u>
we apply the relation below
Mg - T = Ma --- ( 1 )
M = 69kg
g = 9.81
T = 58 * 9.81
a = ? ( magnitude of minimum acceleration)
From equation 1
a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69
= 1.5639 m/s^2
To 2 significant Figures = 1.6 m/s^2
Answer:
first order date and most recent order date
Explanation:
it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019
Answer:
the linear speed of the car is 28.83 m/s
Explanation:
Given;
radius of the car, r = 0.33 m
angular speed of each tire, ω = 13.9 rev/s = 13.9 x 2π = 87.35 rad/s
The linear speed of the car is calculated as;
V = ωr
V = 87.35 rad/s x 0.33 m
V = 28.83 m/s
Therefore, the linear speed of the car is 28.83 m/s
