3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the
1 answer:
Answer:
The wife have to sit at 0.46 L from the middle point of the seesaw.
Explanation:
We need to make a sketch of the seesaw and the loads acting over it.
And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.
A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.
The maximum momentum with respect to the pivot point (0) will be:
Where L/2 is the half of the distance of the seesaw
Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.
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Answer:
q = 4.5 nC
Explanation:
given,
electric field of small charged object, E = 180000 N/C
distance between them, r = 1.5 cm = 0.015 m
using equation of electric field
k = 9 x 10⁹ N.m²/C²
q is the charge of the object
now,
q = 4.5 x 10⁻⁹ C
q = 4.5 nC
the charge on the object is equal to 4.5 nC
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration,
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
Answer:
- the forces on the left hand side is 1.038 kN
- the forces on the right hand side is 1.483 kN
Explanation:
Given the data in the question, as illustrated in the image below;
Length of the scaffold = 3 m
weight of the scaffold = 395 N
Weight of Bob = 805 N and stands 1 m from the left end
weight of washing equipment = 500N and on sits 2 m from the left end
Weight of Joe = 820 N and stand 0.500 m from the right end
so the force on the left cable will be;
=
[ (805 N)( (3-1) m) + ( 395 N )(
) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]
=
[ 1610 + 592.5 + 500 + 410 ]
=
[ 3112.5 ]
= 1037.5 N
= 1.038 kN
Therefore, the forces on the left hand side is 1.038 kN
On the right hand side;
= ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N
= 2520 N - 1037.5 N
= 1482.5 N
= 1.483 kN
Therefore, the forces on the right hand side is 1.483 kN
