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Andrew [12]
2 years ago
14

The length of an aluminum wire is quadrupled and the radius is doubled. By which factor does the resistance change?

Physics
1 answer:
erik [133]2 years ago
7 0
The resistance of a conductive wire is given by:
R= \frac{\rho L}{A}
where
\rho is the material resistivity
L is the wire length
A is the cross-sectional area of the wire

The length of the wire is quadrupled, so if we call L the original length and L' the new length, we can write 
L'=4 L

Similarly, the radius of the wire is doubled (r'=2r), so the new area is
A'= \pi (r')^2 = \pi (2r)^2 = 4 \pi r^2 = 4A

And if we substitute into the equation, we find that the new resistance of the wire is
R'= \frac{\rho L'}{A'}= \frac{\rho (4L)}{4 A'}  =  \frac{\rho L}{A}=R
Therefore, R=R': this means that the resistance of the wire did not change.
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A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box
Georgia [21]

The force exerted on the box is 56 N

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the  force is parallel to the displacement, \theta=0

Solving the equation for F, we find the force exerted on the box:

F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N

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Answer:

There are nine PCNs in the Edmonton area. They work alongside more than 1,100 family doctors in over 330 clinics to provide care for 1.2 million patients. PCN teams include more than 370 nurses, mental health clinicians and other health professionals.

Explanation:

3 0
3 years ago
How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil
mrs_skeptik [129]
The frequency of a simple harmonic oscillator such as a spring-mass system is given by
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m is the mass attached to the spring.

Re-arranging the formula, we get:
m= \frac{k}{4 \pi^2 f^2}
and since we know the constant of the spring:
k=39.5 N/m
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg
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The formula for the acceleration due to gravity is:

a = Gm/r²
where
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m is the mass of planet
r is the radius of planet

So, if they have the same a:

m₁/r₁² = m₂/r₂²
So, if m₁ = m and r₂ = 2r₁,
m/r₁² = m₂/(2r₁)²
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<em>Thus, the answer is D.</em>
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Answer and Explanation:

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