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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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A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 \frac{m}{s}

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Explanation:

Length legs=L=1.1m

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the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

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d).

The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

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