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zhenek [66]
2 years ago
14

A book weighing 1.2 kg stands on the table. We place our hand on the book with a force of 15.3 N. The reaction force of the tabl

e base to the book is: a) 12 N b) 15 N c) 22 N d) 27 N e) 33 N​
Physics
1 answer:
babunello [35]2 years ago
8 0

Answer:

27 N

Explanation:

It will be the force of the book  1.2 kg * 9.81 m/s^2   PLUS the force of your hand  15.3 N    ..... it will be in the OPPOSITE direction

1.2 * 9.81 + 15.3 = ~27.1 N

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When two forces are acting a body in opposite direction their resultant become 6N and when they are acting in the same direction
Norma-Jean [14]

Answer:

9 and 3 N

Explanation:

Forces in the same direction sum up to produce the resultant force;

One force subtract the other will give the resultant force when they are in opposite directions;

Lets say one direction is forwards and the opposite backwards;

We have one force, let's say force A, in the forwards direction and another force, force B, acting in the same (forwards) or opposite (backwards) direction;

If B is acting in the same direction, then the resultant force (in this case) will be as follows:

A + B = 12

If B is acting in the opposite direction, then the resultant force will be as follows:
A - B = 6

Summing the two equations will allow us to solve for A:

A + B + (A - B) = 12 + 6

2A = 18

A = 9

Substitute this into either of the above equations and we can solve for B:
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B = 9 - 6

B = 3

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A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. Th
ipn [44]

Answer:

a) F_{fric} = 692 N

b) F_{applied} = 932 N

Explanation:

a)

According to newton's second law of motion, acceleration of an object is directly proportional to the net force acting on it. When there is no net force force acting on the body, there is no acceleration. A force is a push or a pull, and the net force ΣF is the total force, or sum of the forces exerted on an object  in all directions.

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

F_{applied} = 800 N

Mass = m = 90 kg

acceleration = a = 1.2 m/s²

F_{fric} = ?

800 - F_{fric} = (90)(1.2)

F_{fric} = 692 N

b)

According to newton's second law of motion,

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

If we assume the same friction and acceleration between player's feet and ground as calculated in part a

F_{fric} = 692 N

acceleration = a = 1.2 m/s²

We take the equal mass to the total mass of both the players because when the winning player push losing player backward, he exert force on the ground not only due to his mass but also due to the mass of losing player.

Mass = M = m₁ + m₂ = 110 kg + 90 kg

= 200 kg

F_{applied} = ?

F_{applied} - 692 N = (200)(1.2)

F_{applied} = 692 + 240

F_{applied} = 932 N

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4 years ago
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Answer:

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