Answer:
Power of the man is 132.9 W.
Explanation:
Given that,
Work performed by the man is 2525 J
Time for which the work is performed is 19 second
To find :
Power of the man
Solve :
Let P is the power of the man. The total work done divided by time taken is called the power of an object. Its formula is given by :
or
Therefore, the power of the man is 132.9 W.
Answer:
K = G Mm / 9R
Explanation:
Expression for escape velocity V_e =
Kinetic energy at the surface = 1/2 m V_e ²
= 1/2 x m x 2GM/R
GMm/R
Potential energy at the surface
= - GMm/R
Total energy = 0
At height 9R ( 8R from the surface )
potential energy
= - G Mm / 9R
Kinetic energy = K
Total energy will be zero according to law of conservation of mechanical energy
so
K - G Mm / 9R = 0
K = G Mm / 9R
Answer:
Answer:196 Joules
Explanation:
Hello
Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem
the work is the product of a force applied to a body and the displacement of the body in the direction of this force
assuming that the force goes in the same direction of the displacement, that is upwards
W=F*D (work, force,displacement)
the force necessary to move the object will be
Answer:196 Joules
I hope it helps
This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
