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Delvig [45]
3 years ago
14

(quick help please??)A mechanic uses a mechanical lift to raise a car. The car weighs 10,200 N. The work required to do this was

22,440 J.
What is the distance over which the force was applied?
A. 4.64 J
B. 2.04 J
C. 2.2 m
D. 2.16 10^-4 m
Physics
1 answer:
GuDViN [60]3 years ago
5 0
I believe It would be 2.2m
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A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle
Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

  The mass of the wheel is  m =  6.9  kg

   The radius is  r =  0.69 \  m

    The radius of gyration is  k_G = 0.4\  m

    The angle is  \theta = 47^o

    The force which the massless bar is subjected to F = 22.5 \  N

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that  point A is the  center of rotation.

  Generally the moment of  inertia about A is mathematically represented as

     I_a =  I_G + M* r^2

Here I_G is the moment of inertia about G with respect to the radius of gyration  which is mathematically represented as

    I_G =  M *  k_G

=>I_a = k_G*  M + M* r^2

=>I_a =0.4 * 6.9  + 6.9 * 0.69^2

=>I_a =6.045 \  kg \cdot m^2

Generally the torque experienced by the wheel  is mathematically represented as

       \tau =  F *  cos (47)

=>     \tau =  22.5 *  cos (47)

=>     \tau =  15.34 \ kg \cdot m^2 \cdot s^{-2}

Generally this torque is also mathematically represented as

     \tau = I_a * \alpha

=>   15.34  =  6.045 * \alpha

=>   \alpha =2.538 \  rad/s^2

4 0
3 years ago
Need 3 examples of unbalanced forces
jarptica [38.1K]
1. pulling tug of war
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4 0
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alexandr1967 [171]

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5 0
3 years ago
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Len [333]

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3 0
2 years ago
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Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0
3 years ago
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