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kenny6666 [7]
3 years ago
10

A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient

of sliding friction between the sidewalk and the metal runners of the sled?

Physics
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

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The Gulf Stream is an ocean current that runs from the southern tip of Florida up the eastern coast of the U.S. And ends in the
dmitriy555 [2]

Answer:

The Northern flowing stream will affect the coast of Europe by  making  climate of Western Europe and Northern Europe to be warmer than other areas that are located around the same latitude and this is simply because of the North Atlantic Current

Explanation:

The Northern flowing stream will affect the coast of Europe because the North Atlantic current causes the Gulf stream to be split into two ( 2 ) and the Northern stream which is a warm stream crossing over to Northern Europe. <em>The Gulf stream makes climate of Western Europe and Northern Europe to be warmer than other areas that are located around the same latitude and this is simply because of the North Atlantic Current</em>

7 0
2 years ago
The distance and displacement of a object in motion can be the same (true or false)
Rasek [7]

Aswer:

False, the values ​​of the distance traveled and the displacement only coincide when the trayectorie is a straight line. Otherwise, the distance will always be greater than the offset.

Although these terms are used synonymously in other cases, they are totally different. Since the distance that a mobile travels is the equivalent of the length of its trajectory. Whereas, the displacement will be a vector magnitude.

<u>xXCherryCakeXx</u>.

4 0
2 years ago
What is the length of a spring that has 450J of potential energy and a spring constant of 650N/m?
11111nata11111 [884]

Answer:

Δx = 1.2 m

Explanation:

The CHANGE of spring length) (Δx) can be found using PS = ½kΔx²

Δx = √(2PS/k) = √(2(450)/650) = 1.17669... ≈ 1.2 m

The actual length of the spring is unknown as it varies with material type, construction method, extension or compression, and other variables we have no clue about.

4 0
3 years ago
At 6: 00 am, a motorbike set off from town A to town B at a speed of 40km/h. At the same time, a car set off from town B to town
Keith_Richards [23]

Answer:

One would need to know how far apart the towns are:

T = SA / 40      time it takes for first cyclist to travel S1

T = SB / 60       time it takes for cyclist B to travel distance S2

SA + SB = S     the distance between the towns

SB = 60 / 40 SA = 1.5 SA

SA + 1.5 SA = S

S = 2.5 SA where cyclist travels distance SA

The time will depend on the separation of the towns.

4 0
1 year ago
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0
3 years ago
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