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Julli [10]
3 years ago
5

A uniform 40-N board supports two children weighing 500 N and 350 N. If the support is at the center of the board and the 500-N

child is 1.5 m from its center, How far is the 350-N child from the center?
Physics
1 answer:
serg [7]3 years ago
5 0

Answer:

b= 2.14 m

Explanation:

Given that

Weight of the board ,wt = 40 N

Wight of the first children , wt₁=500 N

Weight of the second children ,wt₂ = 350 N

The distance of the 500 N child from center ,a= 1.5 m

lets take distance of the 350 N child from center = b m

Now by taking the moment about the center of the board

We know that moment = Force x Perpendicular distance from the force

wt₁ x a = wt₂ x b

500 x 1.5 = 350 x b

b= 2.14 m

Therefore the distance of the 350 N weight child from the center is 2.14 m.

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During electrical storms, a bolt of lightning can transfer 10 C of charge in 2.0 µs (the amount of time can vary considerably).
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(a)

distance, d = 1 m

the formula for the magnetic field is given by

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B = 10^{-7}\frac{2\times 5\times 10^{6}}{1}

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4 years ago
In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe
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Answer:

Explanation:

Given

Each student exert a force of F=400 N

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3 years ago
Although the skier has a jacket on, she is still cold. How can her circulatory system help keep her warm?
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6 0
3 years ago
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Calculate the force of a particle with a net charge of 170 coulombs traveling at a speed of 135 meters/second perpendicular to t
amm1812

Answer:

F=1.14N j

Explanation:

The magnitude of the magnetic force over a charge in a constant magnetic field is given by the formula:

|\vec{F}|=|q\vec{v} \ X\ \vec{B}|=qvsin\theta  (|)

In this case v and B vectors are perpendicular between them. Furthermore the direction of the magnetic force is:

-i X k = +j

Finally, by replacing in (1) we obtain:

\vec{F}=(170C)(135\frac{m}{s})(5.0*10^{-5}T)=1.14N\ \hat{j}

hope this helps!

6 0
3 years ago
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