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jeyben [28]
3 years ago
11

Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2

2.4 N. How much combined work is done on the wagon after it has moved 2.50 meters to the right?
Physics
2 answers:
Sidana [21]3 years ago
7 0

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

F_n=F-F'

F_n=37.2-22.4

F_n=14.8\ N

Work done on the wagon is given by the product of net force and displacement. It is given by :

W=F_n\times d

W=14.8\ N\times 2.5\ m

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

Novosadov [1.4K]3 years ago
3 0

Answer:

Explanation:

Rightwards force, F1 = 37.2 N

Leftwards force, F2 = 22.4 N

distance, d = 2.5 m

Net force, F = F1 - F2

F = 37.2 - 22.4

F = 14.8 N rightwards

Work done, W = force x distance

W = 14.8 x 2.5

W = 37 J

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3 years ago
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What is the maximum force that could be applied to anterior cruciate ligament (ACL) if it has a diameter of 4.8 mm and a tensile
vovangra [49]

Answer:

Maximum force, F = 1809.55 N

Explanation:

Given that,

Diameter of the anterior cruciate ligament, d = 4.8 mm

Radius, r = 2.4 mm

The tensile strength of the anterior cruciate ligament, P=100\times 10^6\ N/m^2=10^8\ Pa

We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,

F=P\times A\\\\F=10^8\times \pi (2.4\times 10^{-3})^2\\\\F=1809.55\ N

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4 0
4 years ago
you slide across home plate during baseball game. If you have a mass of 82 kg, and the coefficient of kinetic friction between y
Tju [1.3M]

m = mass of the person = 82 kg

g = acceleration due to gravity acting on the person = 9.8 m/s²

F = normal force by the surface on the person

f = kinetic frictional force acting on the person by the surface

μ = Coefficient of kinetic friction = 0.45

The normal force by the surface in upward direction balances the weight of the person in down direction , hence

F = mg                                          eq-1

kinetic frictional force on the person acting is given as

f = μ F

using eq-1

f = μ mg

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f = (0.45) (82) (9.8)

f = 361.6 N

8 0
3 years ago
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