High task repitition, forceful exertions, repetitive or sustained awkward posture
A. The acceleration during the slide is 6.86 m/s²
B. The time taken to slide until he stops is 1.2 s
<h3>How to determine the force of friction</h3>
- Mass (m) = 81.5 Kg
- Coefficient of friction (μ) = 0.7
- Acceleration due to gravity (g) = 9.8 m/s²
- Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
- Frictional force (F) =?
F = μN
F = 0.7 × 798.7
F = 559.09 N
<h3>A. How to determine the acceleration</h3>
- Mass (m) = 81.5 Kg
- Frictional force (F) = 559.09 N
- Acceleration (a) =?
a = F / m
a = 559.09 / 81.5
a = 6.86 m/s²
<h3>B. How to determine the time </h3>
- Initial velocity (u) = 8.23 m/s
- Final velocity (v) = 0 m/s
- Decceleration (a) = -6.86 m/s²
- Time (t) =?
a = (v – u) / t
t = (v – u) / a
t = (0 – 8.23) / -6.86
t = 1.2 s
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Acceleration is the change in velocity
One of the major limitations of using the ball and stick model for DNA, is that within a single double stranded segment of DNA, one would have to use many many balls to represent atoms that are present in the sugar phosphate backbone, along with all of the main atoms that compose the nitrogenous bases of DNA, we also cannot construct or show the helical form of DNA, by using balls and sticks as well.
Answer:
The frictional force needed to overcome the cart is 4.83N
Explanation:
The frictional force can be obtained using the following formula:

where
is the coefficient of friction = 0.02
R = Normal reaction of the load =
=
= 
Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F = 4.83 N
Hence, the frictional force needed to overcome the cart is 4.83N