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ollegr [7]
3 years ago
8

A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa

gon handle, which is inclined at 35° to the sidewalk as shown below. What is the magnitude of the fore of friction on the wagon?
Physics
1 answer:
Jlenok [28]3 years ago
5 0

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is  

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx =  18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

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A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio
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The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

W=p\Delta V = p(V_f - V_i)

where

p is the pressure of the gas

V_i is the initial volume

V_f is the final volume

For the gas in the cylinder in this problem,

p = 2.00 atm

V_i = 0.250 L

And we also know the work done,

W = 288 J

So we can solve the equation for V_f, the final volume:

V_f = V_i + \frac{W}{p}=0.250 + \frac{288}{2.00}=144.25 L

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3 years ago
Define average velocity.Immersive Reader
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The slope of a graph of position vs time

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3 years ago
Manuel is holding a 5 kg box. How much force us the box exerting on him? in what direction
Free_Kalibri [48]
It will be 49 Newtons of force in the down direction. To find the force in newtons, you multiply the mass (5 kg) by the gravity (which if 9.8). 
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3 years ago
Changes of state
Mrrafil [7]

Answer:

Option A

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4 0
2 years ago
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A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

s=\dfrac{t(u+v)}{2}

u=\dfrac{2s}{t}-v

u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

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