Help me with this PLEASE, I'M being timed

Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long

nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

$R = \dfrac{\rho L }{A}$

where;

A = πr²

$R = \dfrac{\rho L }{\pi r ^2}$

For the shorter cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{ 4 \rho L }{\pi \ D ^2}$

For the longer cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{32\rho L }{\pi \ (4 D) ^2}$

$R = \dfrac{2\rho L }{\pi \ (D) ^2}$

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

$\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}$

$\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}$

$\dfrac{R_s}{R_L} =2$

${R_s}=2{R_L}$

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0

3 years ago

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

3 years ago

(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en

V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0

2 years ago

The distance-time graph for a faster moving object has a smaller slope than the graph for a slower moving object. True or false

allsm [11]

False because faster moving slope would be going up and slower would be going down because its decreasing

6 0

3 years ago

Read 2 more answers

Which of these is most likely to lead directly to a black market?

Sloan [31]

The answer is C.rationing.

<span>♡♡Hope I helped!!! :)♡♡
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7 0

3 years ago

Read 2 more answers