A box of books weighing 328 N moves witha constant velocity across the floor when it is
pushed with a force of 378 N exerted downward at an angle of 25.6◦ below the horizontal.
Find µk between the box and the floor.

Answers

Answer 1

Answer:

Answer:

Approximately $0.694$ , assuming that the floor is level.

Explanation:

Between two surfaces that are moving relative to one another, the coefficient of kinetic friction $\mu_(k)$ is equal to the ratio between friction and normal force.

Since the box in this question is moving at a constant speed, the box would be in a translational equilibrium. Forces on this box should be balanced in both the horizontal component and the vertical component.

The value of $\mu_(k)$ in this question can be found in the following steps:

Decompose the external force into horizontal and vertical components.
Balance forces in the horizontal direction to obtain an expression for the friction on this object.
Balance forces in the vertical direction to find an expression for the normal force on this object.
Divide the magnitude of friction by the magnitude of the normal force to find the coefficient of kinetic friction, $\mu_(k)$ .

At an angle of $\theta = 25.6^(\circ)$ from the horizontal, magnitude of the vertical and horizontal components of the external force would be:

Horizontal component: $F\, \cos(\theta) = (378\; {\rm N})\, \cos(25.6^(\circ))$ .
Vertical component: $F\, \sin(\theta) = (378\; {\rm N})\, \sin(25.6^(\circ))$ .

Assume that the floor is level. Forces on this object in the horizontal direction would include:

Horizontal component of the external force, pointing in the direction of motion.
Friction from the ground, pointing backward opposite to the direction of motion.

Forces on this object are balanced in the horizontal direction. Hence, the magnitude of friction would be equal to that of the horizontal component of the external force:

$(\text{magnitude of friction}) = F\, \cos(\theta)$ .

Forces on this object in the vertical direction would include:

Weight of this object, pointing downward.
Vertical component of the external force, pointing downward.
Normal force from the ground, pointing upward.

Forces on this object in the vertical direction are also balanced. The magnitude of the normal force (pointing upward) should be equal to the sum of the magnitude of the two forces pointing downward:

$\begin{aligned} & (\text{magnitude of normal force}) \n =\; & (\text{magnitude of weight}) + F\, \sin(\theta)\end{aligned}$ .

It is given that the magnitude of the weight of this object is $328\; {\rm N}$ . To find the coefficient of kinetic friction, divide the magnitude of friction by the magnitude of the normal force:

$\begin{aligned}\mu_(k) &= \frac{(\text{magnitude of friction})}{(\text{magnitude of normal force})} \n &= \frac{F\, \cos(\theta)}{(\text{magnitude of weight}) + F\, \sin(\theta)} \n &= \frac{(378\; {\rm N})\, \cos(25.6^(\circ))}{(328\; {\rm N}) + (378\; {\rm N})\, \sin(25.6^(\circ))} \n &\approx 0.694\end{aligned}$ .

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Answers

Final answer:

The question is related to Physics and deals with kinematic equations. With the supplied information, one can calculate elements such as velocity or applied force in the jump.

Explanation:

The subject of the question pertains to the field of Physics, specifically the area of kinematic equations which deal with the motion of objects. The provided information in the question pertains to the rise of a person's body during a jump. Given the average height of 60cm that a person typically attains and the approximate rise of the body from the knees up being 50cm, these figures can be used in a Physics context to determine different factors of the jump such as velocity or force applied.

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On a highway curve with radius 30 m, the maximum force of static friction (centripetal force) that can act on a 1,423-kg car going around the curve is 8,127 n. what speed limit should be posted for the curve so that cars can negotiate it safely?

Answers

To determine the speed limit for the curve, we can use the centripetal force equation. The centripetal force (Fc) is given by the equation Fc = (m * v^2) / r, where m is the mass of the car, v is the velocity, and r is the radius of the curve.

In this case, the maximum force of static friction acting as the centripetal force is given as 8,127 N, the mass of the car is 1,423 kg, and the radius of the curve is 30 m.

We can rearrange the equation to solve for the velocity (v):

v = sqrt((Fc * r) / m)

Now, let's substitute the given values into the equation:

v = sqrt((8,127 N * 30 m) / 1,423 kg)

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Therefore, to ensure that cars can negotiate the curve safely, the speed limit should be posted at approximately 13.57 m/s (or about 49 km/h). It's important to note that this is only a rough estimate, and other factors such as road conditions and driver behavior should also be taken into consideration when setting speed limits.

Final answer:

The speed at which a car can safely navigate a given curve is determined by the equation for centripetal force and the maximum static friction that prevents the car from slipping. Calculating this speed using the given force of static friction (8,127 N), mass of the car (1,423 kg) and the radius of the curve (30 m), we get a result of roughly 22.6 m/s or 81.4 km/h.

Explanation:

The problem involves finding the speed at which a car can safely navigate a highway curve without slipping, given a set radius and maximum force of static friction. This is a physics scenario involving centripetal force and friction. Centripetal force is the net force causing circular motion and static friction is the friction that acts to prevent the car from slipping off the road.

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Answers

Earthquake S-wave, radio waves, X Ray's, Visible light are examples of transverse waves.

What are waves?

Wave can be defined as any form of disturbance which is transmitted through a medium and energy is transferred from one medium to another thereby causing displacement of the medium.

A Transverse wave is a type of wave in which the wave there is a vibration of the wave medium which is as a result of the wave movement which now become perpendicular to the direction of the wave.

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Answers

Answer:

D or 49.7°

Explanation:

You are given the equation and all the information you need, so you simply need to understand what the question asks for and answer appropriately. Notice that the light wave travels from the water to air. This means that water should be labelled with a "1" as it comes prior to air, which should be labeled "2". Thus, all you need to do, is plug and chug:

$\theta_2 = sin^(-1)((n_1sin(\theta_1))/(n_2)) = sin^(-1)(((1.33)sin(35))/(1))$

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And, therefore, your answer is D, 49.7°.

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Answers

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