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4 years ago

How can the IMA of a first- class lever be increased?

2 answers:

4 0

IMA = Ideal Mechanical Advantage

First class lever = > F1 * x2 = F2 * x1

Where F1 is the force applied to beat F2. The distance from F1 and the pivot is x1 and the distance from F2 and the pivot is x2

=> F1/F2 = x1 /x2

IMA = F1/F2 = x1/x2

Now you can see the effects of changing F1, F2, x1 and x2.

If you decrease the lengt X1 between the applied effort (F1) and the pivot, IMA decreases.

If you increase the length X1 between the applied effort (F1) and the pivot, IMA increases.

If you decrease the applied effort (F1) and increase the distance between it and the pivot (X1) the new IMA may incrase or decrase depending on the ratio of the changes.

If you decrease the applied effort (F1) and decrease the distance between it and the pivot (X1) IMA will decrease.

Answer: Increase the length between the applied effort and the pivot.

baherus [9]4 years ago

4 0

B.increase the length between the applied effort and the privot

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Efficiency is completing a task as fast a possible the the least amount of effort.

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3 years ago

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Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.61 kg mass to the

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Given: Mass m = 0.61 Kg; Displacement x = 4.0 cm or x = 0.04 m

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k = F/x

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4 years ago

A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa

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Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

given,

weight of the person = 625 N

weight of the bike = 98 N

Pressure on each Tyre = 7.60 x 10⁵ Pa

Area of contact on each Tyre = ?

total weight of the system = 625 + 98

= 723 N

Let F be the force on both the Tyre

F + F = W

2 F = 723

F = 361.5 N

F = P A

$A = \dfrac{F}{P}$

$A = \dfrac{361.5}{7.60 \times 10^5}$

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4 years ago

When an object is lifted 6 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 12

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When the object is lifted 6 meters of the ground , then lifted to 12 meters, it is lifted twice as high ( 6 x 2 = 12).

This means the potential energy is also doubled.

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3 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

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Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

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4 years ago