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Andre45 [30]
3 years ago
15

A gold wire has a cross-sectional area of 1.0 cm^2 and a resistivity of 2.8 × 10^-8 Ω ∙ m at 20°C. How long would it have to be

to have a resistance of 0.001 Ω?
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0

Answer:

Length, l = 3.57 meters

Explanation:

It is given that,

Area of cross-section of gold wire, A=1\ cm^2=0.0001\ m^2

Resistivity of gold wire, \rho=2.8\times 10^{-8}\ \Omega-m

Resistance, R = 0.001 ohms

Resistance in terms of length and area is given by :

R=\rho \dfrac{l}{A}

l=\dfrac{RA}{\rho}

l=\dfrac{0.001\times 0.0001}{2.8\times 10^{-8}}

l = 3.57 meters

So, the length of the wire is 3.57 meters. Hence, this is the required solution.

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A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
When making a DNA Fingerprint, the job of the restriction enzyme is to..
pogonyaev

Answer:

A isolate the DNA From a sample of cells

4 0
2 years ago
A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from
krek1111 [17]

Answer:

E=35921.96N/C

Explanation:

From the question we are told that:

Radius r=0.321mm

Charge Density \mu=0.100

Distance d= 5.00 cm

Generally the equation for electric field is mathematically given by

E=\frac{mu}{2\pi E_0r}

E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}

E=35921.96N/C

4 0
3 years ago
You are designing a simple elevator system for an old warehouse that is being converted to loft apartments. A 22,500-N elevator
ICE Princess25 [194]

Answer:

A) mass = 3121.58 kg

B) tension = 25940.37 N

C) tension = 25940.37 N (tension on both sides will be the same)

Explanation:

Weight of elevator = 22500 N

Distance = 6.75 m

Time = 3 sec

Since it started from rest, initial speed is zero.

Using Newton's equation of motion we have,

S = ut + 0.5at^2

S = distance covered = 6.75 m

t = time = 3 s

a = acceleration upwards

u = initial velocity = 0

Substituting values, we have,

6.75 = 0(3) + (0.5 x a x 3^2)

6.75 = 4.5a

a = 6.75/4.5 = 1.5 m/s^2 (acceleration of the elevator upwards)

Mass of the elevator = Weight/g

Where g = acceleration due to gravity 9.81 m/s

Mass = 22500/9.81 = 2293.58 kg

From the image below we solve from

T - 22500 = ma

T - 22500 = 2293.58 x 1.5

T - 22500 = 3440.37

T = 3440.37 + 22500 = 25940.37 N (this is the tension on the rope)

On the other side,

mg - T = ma

9.81m - 25940.37 = 1.5m

(9.81 - 1.5)m = 25940.37

8.31m = 25940.37

m = 3121.58 kg (mass of counter weight)

See image below

8 0
3 years ago
The velocity of a mass moving in a circular path is..
Alenkasestr [34]
Constantly changing.
5 0
3 years ago
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