Given Data: Diameter 'd' = 30 cm = 0.3 m Lifting Weight 'W' = mg = 2000*9.81 N = 19,620 N Calculations: Area of the lift 'A' = <span>pi\over4*d^2=pi\over4*0.3^2=0.07 m^2
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Answer:
option A
Explanation:
given,
Force = F
angle = θ
weight on suitcase = mg
distance = d
constant velocity so, acceleration a = 0
coefficient of friction = µ
Work done = ?
Work done is equal to force into displacement.
Friction act opposite to the force acting so, work done by frictional force will be negative.
frictional force will act into horizontal direction opposite to force.
here displacement is equal to d
now,
W = -F d cos θ
Hence,the correct answer is option A
Answer:
4.0 m/s
Explanation:
In the first part of the run, the athlete runs a distance of

at a speed of

So, the time he/she takes is

In the second part of the run, the athlete covers an additional distance of

with a speed

So, the time taken in this second part is

So, the total distance covered is
d = 300 m + 300 m = 600 m
And the total time taken
t = 100 s + 50 s = 150 s
Therefore, the average speed for the entire trip is
