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MaRussiya [10]
2 years ago
12

A 22.8 kg rocking chair begins to slide across the carpet when the push reaches 57.0 N. What is the coefficient of static fricti

on?​

Physics
1 answer:
vfiekz [6]2 years ago
7 0

Answer:

0.255

Explanation:

The following data were obtained from the question:

Force (F) = 57 N

Mass (m) = 22.8 Kg

Coefficient of static friction (µ) =...?

Next, we shall determine the normal reaction (R). This is illustrated below:

Mass (m) = 22.8 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (R) =?

R = mg

R = 22.8 x 9.8

R = 223.44 N

Finally, we can obtain the coefficient of static friction (µ) as follow:

Force (F) = 57 N

Normal reaction (R) = 223.44 N

Coefficient of static friction (µ) =...?

F = µR

57 = µ x 223.44

Divide both side by 223.44

µ = 57/223.44

µ = 0.255

Therefore, the coefficient of static friction (µ) is 0.255.

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Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam
zhannawk [14.2K]

The velocity of Ned as measured by Pam is the interpretation of v.

Answer: Option D

<u>Explanation:</u>

According to question, we know that this is an issue depending on the logical and translation of the factors. From the measured information taken what is gathered by the two people is communicated and we have given as:

The Ned reference framework : (x, t)  

The Pam reference framework :  \left(x^{\prime}, t^{\prime}\right)

From the reference framework, we realize that ν is the speed of Pam (the other reference framework) as estimation by Ned.  

At that point, v^{\prime} is the speed of Ned (from the other arrangement of the reference) as estimation by Pam.

3 0
2 years ago
How far must 5N force pull a 50g toy car if 30J of energy are transferred?​
Alborosie

Answer: 6 m

Explanation:

30 = 5 * d

d = 30/5

d = 6 m

7 0
1 year ago
A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are
Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

4 0
2 years ago
You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same p
Schach [20]
Velocity = (displacement) / (time)

Displacement = straight-line distance between start-point and end-point

If you stop at the same point you started from, then
your displacement for the trip is zero, and your average
velocity is also zero.

5 0
1 year ago
Read 2 more answers
A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di
Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

F . t = m(v_f - v_i)

so here we know that object is at rest initially so we have

80 (12) = 80( v_f - 0)

v_f = 12 m/s

Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

F = 80 - \frac{40}{3} t

now again by same equation we have

\int F .dt = m(v_f - v_i)

\int (80 - (40/3)t) dt = 80(v_f - 12)

80 t - \frac{40t^2}{6} = 80(v_f - 12)

put t = 6 s

480 - 240 = 80(v_f - 12)

v_f = 12 + 3

v_f = 15 m/s

6 0
1 year ago
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