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3 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]3 years ago

4 0

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

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Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago

ANSWER PLS NEED ASAP AS FAST AS HUMANLY POSSIBLE

Soloha48 [4]

Answer:

Should be B

Explanation:

As the light enters the prism at an angle, the glass causes the light to travel slower. The speed is affected by the frequency of the light. So each frequency is slowed by a different amount, causing the colors to separate.

Hope this helps!

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3 years ago

In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied

Lesechka [4]

K = 1/2 m x v^2

m = mass on the cart

V = velocity imparted to the cart

KA = 1/2 mA x vA^2.......................(1)

KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

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Option A is correct

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3 years ago

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Semmy [17]

Answer:

480

Explanation:

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120÷0.25

=480

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blondinia [14]

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