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dmitriy555 [2]
3 years ago
7

w strong are you? Could you turn a moving car with just your own strength? No way! If a normal driver inputs about 50 N of force

to turn a car, the output force is 3,750 Newtons! What is the MA of the steering wheel?
Physics
1 answer:
malfutka [58]3 years ago
3 0

The mechanical advantage of a simple machine is the measure of its amplified force gain.

The mechanical advantage  is defined as the force amplified by a machine to the force required to generate such output.

Mathematically\ mechanical\ advantage\ MA=\frac{F_{o}} {F_{i}}

F_{o} \ and\ F_{i} are the amplified force and applied force. We may also consider them as output and input force.          


In the given question, the force given to the steering wheel is 50 N.

The output force produced by the steering wheel is 3750 N.

Hence the mechanical advantage will be-

                               MA=\frac{F_{o}} {F_{i}}

                                       =\frac{3750\ N}{50\ N}

                                       =75      [ans]

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JulijaS [17]
Total amount of energy would remain constant according to law of conservation of energy. i.e., 50 Joules

In short, Your Answer would be Option C) <span>50 Joules because as energy converts from one form to another, it cannot be created or destroyed during the conversion.
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7 0
3 years ago
Read 2 more answers
A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal
dsp73

Answer:

v\approx 8.570\,\frac{m}{s}

Explanation:

The equation of equlibrium for the box is:

\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a

The formula for the acceleration, given in \frac{m}{s^{2}}, is:

a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}

Velocity can be derived from the following definition of acceleration:

a = v\cdot \frac{dv}{dx}

v\, dv = a\, dx

\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx

\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg}  \int\limits^{17\,m}_{0\,m}\, dx  - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx

\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}

v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}]  }

The speed after the box has travelled 17 meters is:

v\approx 8.570\,\frac{m}{s}

3 0
3 years ago
Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right c
masya89 [10]

Answer:

vₓ = xg/2y

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In this question, let us  find the time it takes for the ball on the right that has zero initial velocity to reach the ground.

By newton equation of motion we know that

y = v₀ t - ½ g t²

t = 2y / g

This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance

vₓ = x/t

vₓ = xg/2y

vₓ = xg/2y

Where we assume that x and y are known.

7 0
3 years ago
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Answer:

d

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Explanation:

8 0
3 years ago
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Betty (mass 40 kg), standing on slippery ice, catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Show that the
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Answer:

v = 2.18m/s

Explanation:

In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.

Then you have:

Mv_{1o}+mv_{2o}=(M+m)v        (1)

M: mass Betty = 40kg

m: mass of the dog = 15kg

v1o: initial speed of Betty = 3.0m/s

v2o: initial speed of the dog = 0 m/s

v: speed of both Betty and her dog = ?

You solve the equation (1) for v:

v=\frac{Mv_{1o}+mv_{2o}}{M+m}=\frac{(40kg)(3.0m/s)+(15kg)(0m/s)}{40kg+15kg}\\\\v=2.18m/s

The speed fo both Betty and her dog is 2.18m/s

7 0
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