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Natalija [7]
3 years ago
11

How do you find the mechanical advantage of a wheel and axle

Physics
1 answer:
alexandr1967 [171]3 years ago
7 0

Answer:  <em>In order to calculate the mechanical advantage of a wheel and axel, you can divide the radius of the wheel by the radius of the axle. If a wheel's radius is 60cm and its axle is 30cm, what is the mechanical advantage of the wheel and axle? The output force of the wheel and axle is 2 times greater than the input force.</em>

<em />

<h2></h2>

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Can I also get help on this??
Xelga [282]

Answer:

25

Explanation:

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3 years ago
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Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s
g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

v_{x} =  v cos(∅)

v_{y} =  v sin(∅)

We know that for a projectile motion,

t =\frac{2vsin(∅)}{g}

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = v_{x}×t = vcos(∅)×\frac{2vsin(∅)}{g} = \frac{v^{2}sin(2∅) }{g}.

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = \frac{v^{2}sin(2α) }{g} = \frac{v^{2}sin(2[90-β]) }{g} =\frac{v^{2}sin(180-2β) }{g} = \frac{v^{2}sin(2β) }{g} .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

     β = 90-25 = 65°

∴ The required angle is 65°.

5 0
3 years ago
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Additionally, Jet engines have two openings (an intake and an exhaust nozzle). Rocket engines only have one opening (an exhaust nozzle).

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a point charge q1 is held stationary at the origin. a second charge q2 is placed at point a, and the electric potential energy o
Gnoma [55]

The electric potential energy of the pair of charges when the second charge is at point b is U at b = +7.3×10⁻⁸J.

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Here the particle should be moved from the point at the time when the potential energy should be U a

The U b should be the change in the potential energy

W is work done.

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U b = U a - (W at a )+(W at b)

So,

= 5.4 × 10⁻⁸J - ( -1.9×10⁻⁸ J)

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USPshnik [31]

Answer:

The value t = 1.995 \  s

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From the question we are told that

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Here u  is the velocity of the ball at maximum height before it start falling and the value is  0 m/s

So

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3 years ago
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