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ollegr [7]
2 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]2 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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60 mph east...........

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When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscil
marishachu [46]

Answer:

a) 40,000 N/m

b) f = 6.37 Hz

c) v = 4,8 m/s

Explanation:

part a)

First in order to estimate the spring constant k, we need to know the expression or formula to use in this case:

k = ΔF / Δx

Where:

ΔF: force that the men puts in the car, in this case, the weight.

Δx: the sinking of the car, which is 2 cm or 0.02 m.

With this data, and knowing that there are four mens, replace the data in the above formula:

W = 80 * 10 = 800 N

This is the weight for 1 man, so the 4 men together would be:

W = 800 * 4 = 3200 N

So, replacing this data in the formula:

k = 3200 / 0.02 = 160,000 N/m

This means that one spring will be:

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b) An axle and two wheels has a mass of 50 kg, so we can assume they have a parallel connection to the car. If this is true, then:

k^n = 2k

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Replacing the data:

wo = √2 * 40,000 / 50 = 40 rad/s

And the frequency:

f = wo/2π

f = 40 / 2π = 6.37 Hz

c) finally for the speed, we have the time and the distance, so:

V = x * t

The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:

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5 0
2 years ago
Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo
lisabon 2012 [21]

Answer:

El peso del cartel es 397,97 N

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La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

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El peso del signo = 397,97 N.

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the dimensions multiply like algebraic quantities. 
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4 0
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