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3 years ago

The lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?

Physics

1 answer:

DaniilM [7]3 years ago

3 0

Answer:

R = 0.0015Ω

Explanation:

The formula for calculating the resistivity of a material is expressed as;

ρ = RA/l

R is the resistance

ρ is the resistivity

A is the area of the wire

l is the length of the wire

Given

l = 85cm = 0.85m

A = πr²

A = 3.14*0.0018²

A = 0.0000101736m²

ρ = 1.75 × 10-8Ωm.

Substitute into the formula

1.75 × 10-8 = 0.0000101736R/0.85

1.4875× 10-8 = 0.0000101736R

R = 1.4875× 10-8/0.0000101736

R = 0.0015Ω

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A two slit pattern is viewed on a screen 1.00m from the slits if the two third-order minima are 22.0 cm apart what is the width

Bingel [31]

Answer:

4.4 cm

Explanation:

Given:

Distance of the screen from the slit, D = 1 m

Distance between two third order interference minimas, x = 22 cm

Let's say, minima occurs at:

$x_n = (n + \frac{1}{2}) \frac{wL}{d}$

We have:

$2x_2 = 2(2 + \frac{1}{2}) * \frac{w*22}{d}$

Calculating further for the width of the central bright fringe, we have:

$\frac{w}{d} = \frac{22}{5}$

= 4.4 cm

Note: w in representswavelength

8 0

3 years ago

A uniform rod of length L, mass M, is suspended by two thin strings. Which of the following statements is true

Fantom [35]

Answer:

(d) not enough info

Explanation:

because it doesn't specify where the strings are attached

if it was the two ends of the rod then T1 would be equal to T2

6 0

3 years ago

The conducting path between the right hand and the left hand can be modeled as a 9.0-cm-diameter, 140-cm-long cylinder. The aver

Crank

Answer:

The potential difference is 121.069 V

Solution:

As per the question:

Diameter of the cylinder, d = 9.0 cm = 0.09 m

Length of the cylinder, l = 40 cm = 1.4 m

Average Resistivity, $\rho = 5.5\ \Omega-m$

Current, I = 100 mA = 0.1 A

Now,

To calculate the potential difference between the hands:

Cross- sectional Area of the Cylinder, A = $\pi (\frac{d}{2})^{2} = 6.36\times 10^{- 3}\ m^{2}$

Resistivity is given by:

$\rho = R\frac{A}{l}$

$R = \rho \frac{l}{A}$

$R = 5.5\times \frac{1.4}{6.36\times 10^{- 3}} = 1210.69\ \Omega$

Now, using Ohm's Law:

V = IR

$V = 0.1\times 1210.69 = 121.069\ V$

4 0

4 years ago

Find the acceleration that can result from a net force of 13 N exerted on a 3.6-kg cart. (Note: The unit N/kg is equivalent to m

STatiana [176]

Answer:

a = 3.61[m/s^2]

Explanation:

To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.

In this case we have:

$F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]$

7 0

3 years ago

You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

3 years ago