The force applied by the man is 60 N
Explanation:
We can solve this problem by applying Newton's second law, which states that:
(1)
where
is the net force acting on the child+cart
m is the mass of the child+cart system
a is their acceleration
In this problem, we have:
m = 30.0 kg is the mass
And there are two forces acting on the child+cart system:
- The forward force of pushing, F
- The force resisting the cart motion, R = 15.0 N
Therefore we can write the net force as
where R is negative since its direction is opposite to the motion
So eq.(1) can be rewritten as
And solving for F,
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Answer:
The centripetal force acting on the skater is <u>48.32 N.</u>
Explanation:
Given:
Radius of circular track is, 
Tangential speed of the skater is, 
Mass of the skater is, 
We are asked to find the centripetal force acting on the skater.
We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.
Centripetal force acting on the skater is given as:
Now, plug in the given values of the known quantities and solve for centripetal force, 
. This gives,
Therefore, the centripetal force acting on the skater is 48.32 N.
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
Answer:
Expression of work done is
Work done to move the sled is given as 1.94 J
Explanation:
As we know that the formula of work done is given as
here we know that
F = 6 N
d = 0.4 m
so we will have
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ
= σ
( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23
