A dynamics cart with a friction pad is placed at the top of an inclined track and released fromrest. The cart accelerates down the incline at a rate of 0.60 m/s2. If the track is angled at 10degrees above the horizontal, determine the coefficient of kinetic friction between the cart andthe track.

Answers

Answer 1
Answer:

Answer:

0.114

Explanation:

There are two forces acting on the cart in the direction along the cart:

- The component of the gravitational force in the direction parallel to the ramp, mg sin \theta, down along the ramp

- The force of friction, \mu N, up along the ramp

So the equation of motion along this direction is:

mg sin \theta - \mu N = ma (1)

where

m is the mass of the cart

g=9.8 m/s^2 is the acceleration due to gravity

\theta=10^(\circ) is the angle of the ramp

\mu is the coefficient of kinetic friction

N is the normal force exerted by the ramp on the cart

a=0.60 m/s^2 is the acceleration of the cart

The normal force can be found from the equation of the forces along the direction perpendicular to the ramp; in fact, the normal force is balanced by the component of the weight perpendicular to the ramp, so we have:

N-mg cos \theta = 0

From which we get:

N=mg cos \theta

Substituting into (1),

mg sin \theta - \mu mg cos \theta = ma

And solving for \mu, we find the coefficient of friction:

\mu = (g sin \theta - a)/(g cos \theta)=((9.8)(sin10^(\circ))-0.60)/((9.8)(cos 10^(\circ))=0.114


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Answers

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Answers

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Compare and contrast, in detail, at least two types of pluralism. Do you think the United States is becoming more pluralistic? Why or why not? How do ethnic enclaves relate to pluralism in the US?

Answers

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b. Yes, the United States is becoming more pluralistic

c. Ethnic enclaves can be both a positive and negative aspect of pluralism in the US.

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b. The United States is becoming more pluralistic, as the population becomes more diverse and the importance of multiculturalism and political pluralism is increasingly recognized. The country is home to many different ethnic and cultural groups, and these groups are increasingly visible in politics and society. This is evident in the increasing number of ethnic and cultural enclaves, which are areas where different ethnic groups live in close proximity to one another.

c. Ethnic enclaves can be both a positive and negative aspect of pluralism in the US. On the one hand, they allow individuals to maintain their cultural traditions and create supportive communities. On the other hand, they can also lead to segregation and exclusion, as different groups may be reluctant to interact with one another. In order to promote a positive form of pluralism, it is important to encourage interaction and understanding between different groups, while also respecting and celebrating their distinct cultural identities.

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A string is attached to the rear-view mirror of a car. A ball is hanging at the other end of the string. The car is driving around in a circle, at a constant speed. Which of the following lists gives all of the forces directly acting on the ball?a. tension
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Answers

Final answer:

The forces directly acting on the ball hanging from a rear-view mirror while a car drives in a circle are tension, gravity, and the centripetal force.

Explanation:

The correct answer is c. tension, gravity, and the centripetal force.

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A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?

Answers

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1  (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

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Therefore,

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where k = Constant of proportionality.

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Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

A hot-air balloon is descending at a rate of 2.3 m>s when a pas- senger drops a camera. If the camera is 41 m above the ground when it is dropped, (a) how much time does it take for the cam- era to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.

Answers

(a) It takes approximately 2.8956 seconds for the camera to reach the ground.

(b) The velocity of the camera just before it lands is approximately -28.375 m/s (downward).

We have,

Given information:

Initialheight (h₀) = 41 m (above the ground)

Descentrate of the hot-air balloon = -2.3 m/s (negative because it's descending)

We can use the kinematicequations to solve for the time it takes for the camera to reach the ground and its velocity just before landing.

(a)

To find the time it takes for the camera to reach the ground, we can use the following kinematic equation:

h = h₀ + (v₀)t + (1/2)at²

Where:

h = final height (0 m, as it reaches the ground)

h₀ = initial height (41 m)

v₀ = initial velocity (0 m/s, as the camera is dropped)

a = acceleration (acceleration due to gravity, approximately -9.8 m/s²)

t = time (what we're solving for)

Plugging in the values:

0 = 41 + (0)t + (1/2)(-9.8)t²

Simplifying:

-4.9t² = 41

Divide by -4.9:

t² = -41 / -4.9

t² = 8.36734694

Taking the square root:

t = √8.36734694

t ≈ 2.8956 seconds

(b)

To find the velocity just before the camera lands, we can use the following kinematic equation:

v = v₀ + at

Where:

v = final velocity (what we're solving for)

v₀ = initial velocity (0 m/s)

a = acceleration (acceleration due to gravity, -9.8 m/s²)

t = time (2.8956 seconds, calculated in part a)

Plugging in the values:

v = 0 + (-9.8)(2.8956)

v ≈ -28.375 m/s

The negativesign indicates that the velocity is directed downward, which is consistent with the descending motion.

Thus,

(a) It takes approximately 2.8956 seconds for the camera to reach the ground.

(b) The velocity of the camera just before it lands is approximately -28.375 m/s (downward).

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