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Murrr4er [49]
3 years ago
7

A 120-kg refrigerator, 2.00 m tall and 85.0 cm wide, has its center of mass at its geometrical center. You are attempting to sli

de it along the floor by pushing horizontally on the side of the refrigerator. The coefficient of static friction between the floor and the refrigerator is 0.300. Depending on where you push, the refrigerator may start to tip over before it starts to slide along the floor. What is the highest distance above the floor that you can push the refrigerator so that it won't tip before it begins to slide
Physics
1 answer:
horsena [70]3 years ago
8 0

Answer:

Following are the response to the given question:

Explanation:

To address this problem, the notions of friction and torque in the kinematic equations of motion have to be applied.

The friction resistance is defined by

F=\mu mg

Here seem to be our values.

\mu=0.3\\\\m= 120\ kg \\\\g=9.8\ \frac{m}{s^2} \\\\

F=0.3 \times 120 \times 9.8= 36 \times 9.8= 352.8 \ N

Take the brain's mid-size weight halfway to the floor, i.e. d = \frac{0.85}{2} = 0.425 \ m. The torque around the bottom of the cooler should be zero to reach the maximum range.

F \times x= mg \times d\\\\ \text{Re-set for x}\\\\ x=\frac{mg \times d}{F}= \frac{mg \times d}{ \mu m g} =\frac{d}{\mu}=\frac{0.425}{0.3}=1.42 \m

Then we may say that distance before turning is 1.42m.

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Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache
HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

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