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zheka24 [161]
3 years ago
14

Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force

to provide a greater output force. In other words, we can lift a load without having to lift the full weight of the load. Prove using a mathematical relation that Class II levers provide mechanical advantage. You will need to use the rotational equilibrium condition. It will be helpful to draw a picture indicating the movement of the load. Also, to simplify we can assume the force is still perpendicular to by considering the case when the load just starts to move.

Physics
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

The solution and the explanation are in the Explanation section.

Explanation:

According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:

TA = W * (a * cosθ)

The torque due to effort at C point is:

TC = E * (b * cosθ)

The net torque is equal to 0, we have:

Tnet = 0

W * (a * cosθ) - E * (b * cosθ) = 0

E=W\frac{a}{b}

From the figure, you can observe that a/b < 1, thus E < W

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Answer:

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Explanation:

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3 years ago
a bends each wavelength of white light slightly differently so that each wavelength color comes out separated forming a rainbow
lianna [129]
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3 0
2 years ago
A mechanical wave starts when matter is disturbed by a source of ______.
Gwar [14]

Answer:

MATERIAL MEDIUM

Explanation:

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5 0
2 years ago
A special triangular frame is being made for a piece of artwork. The the base of the triangular frame must be 90 cm. If the area
Mashutka [201]

Answer:

Explanation:

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Let the height is h.

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5 0
3 years ago
Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0
3 years ago
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