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3 years ago

15

Wire 1 has a resistance R1. Wire 2 is made of the same material, with length 1/2 as long and diameter 1/4. What is the resistanc

e ratio R2/R1?

1 answer:

hodyreva [135]3 years ago

5 0

-- reduce the length of a wire to 1/2 . . . cut the resistance in half

-- reduce the diameter to 1/4 . . . reduce the cross-section area by (1/4²) . . . increase the resistance by 16x .

-- R2 = (R1) · (1/2) · (16) = 8 · R1

<em>-- R2 / R1 = 8</em>

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A large box has a mass of 500kg and the coefficient of static friction for the box and the floor is 0.45, and the coefficient of

Mumz [18]

A) The friction force while the box is stationary is (the coefficient of static friction)*(the normal force). In this case, the normal force is equal to the gravitational force, or the weight. To move the box, we need a minimum horizontal force that is equal to the friction force. The weight is (500 kg)*(9.81 m/s^2)= 4905 N. So, (0.45)*(4905 N) = 2207.25 N.

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3 years ago

A stone thrown off a bridge 10m above a river has an initial velocity of 8ms at an angle of 25 degrees above the horizontal. Wha

Darina [25.2K]

Explanation:

It is given that,

Height above which the stone was thrown, h = 10 m

Initial velocity of the stone, u = 8 m/s

Angle above the horizontal, $\theta=25^{\circ}$

The horizontal component of velocity is, $u_x=v\ cos\theta=8\ cos(25)=7.25\ m/s$

The vertical component of velocity is, $u_y=v\ sin\theta=8\ cos(25)=3.38\ m/s$

Let t is the time of flight in vertical motion. The second equation of motion is :

$h=u_yt-\dfrac{1}{2}gt^2$

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t = 0.34 seconds

Let s is the range of the stone. It can be calculated as :

$s=u_x\ cos\theta \times t$

$s=7.25\times 0.34$

s = 2.46 meters

So, the range of the stone is 2.46 meters. Hence, this is the required solution.

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3 years ago