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Alexxx
3 months ago
14

A 120-kg refrigerator that is 2.0 m tall and 85 cm wide has its center of mass at its geometrical center. You are attempting to

slide 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 0.30. 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 will not tip before it begins to slide?
A.1.2 m

B.1.0 m

C.1.6 m

D. 1.4 m

E. 0.71 m
Physics
1 answer:
kicyunya [3.2K]3 months ago
6 0

To tackle this problem, it's essential to utilize the principles concerning the Force of Friction and Torque as defined by kinematic motion equations.

The force of friction is defined as

F= \mu mg

We have our values as such,[

\mu=0.3

m=120kg

g=9.8m/s^2

Performing substitution,

F=0.30*120*9.8 = 352.8N

Consider the center of mass of the refrigerator is halfway its height, which is d = 0.85 / 2 = 0.425 m. The torque about the farther lower corner of the refrigerator must be zero to achieve the maximum pulling distance, hence

F*x = mg*d

Re-arranging this for x,

x= \frac{mg*d}{F}

x= \frac{mg*d}{\mu mg}

x= \frac{d}{\mu}

x= \frac{0.425}{0.3}

x = 1.42m

Consequently, we find that 1.42 m is the maximum height at which the refrigerator can be pushed without tipping.

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Answer:

The correct choice is C: points 1, 4, and 5 are equal, followed by 2 and 3 being equal.

Explanation:

Here’s the breakdown:

The electric field from the positive sheets E₁ = б/2E₀

E₂ is from the negative sheet = -б/2E₀

At points 1, 4, and 5, the electric fields created by the sheets oppose each other.

At point 1, the total field is calculated as -E₁ + E₂ = 0.

Similarly, at point A, the total field results in -E₁ - E₂ = 0.

However, at any point in between the plates, the electric field is directed consistently in one way.

At points 2 and 3, the field is directed to the right.

Thus, we have:

E net = E₁ + E₂

= б/2E₀ + -б/2E₀

=б/E₀

Note: Please refer to the attached document for the full question accompanying this solution.

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3 months ago
A system expands from a volume of 1.00 l to 2.00 l against a constant external pressure of 1.00 atm. what is the work (w) done b
Keith_Richards [3271]
The amount of work performed by a system at consistent pressure is defined by the following equation:
W=p \Delta V = p (V_f - V_i)
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W=p (V_f -V_i)=(1.00 atm)(2.00 L-1.00 L)=1.00 L\cdot atm

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3 months ago
Read 2 more answers
A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a
inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

C=\frac{\epsilon_0A}{d}

In this equation, \epsilon_0 refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

C \alpha \frac{1}{d}.......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

\frac{C_1}{C_2} =\frac{d_2}{d_1}......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

E_1=\frac{Q^2}{2C_1}......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

E_2=\frac{Q^2}{2C_2}......(4)

By dividing equation (4) by (3),

\frac{E_2}{E_1} =\frac{C_1}{C_2}

According to equation (2),

\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}

This results in a 3.5 fold increase in energy.

\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

E_1=\frac{1}{2} C_1V^2......(5)

The final energy when the plate separation transitions to d₂ can be written as:

E_2=\frac{1}{2} C_2V^2.....(6)

Referencing equations (5) and (6)

\frac{E_2}{E_1} =\frac{C_2}{C_1}

From equation (2),

\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}

Thus, in this particular scenario,

\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm





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