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slava [35]
3 years ago
5

How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight

battery?
Physics
1 answer:
Natalka [10]3 years ago
4 0
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
You might be interested in
Given that coal used by electric power plants has a heating value of 27.5 million btus metric ton (25 million btus per ton), det
fredd [130]

Answer:

• 36.4 kg of coal.

• 80 pounds of coal.

Explanation:

Using proportionality constant,

Mass of coal = 1,000,000/27,500,000 btus/metric ton

= 0.0364 metric tons of coal

Mass of coal = 1,000,000/25,000,000 btus/ton

= 0.04 tons of coal.

Converting metric tons to kilogram,

1 metric ton = 1000kg,

0.0364 metric ton;

= 36.4 kg of coal.

Converting tons to pounds,

1 ton = 2000 pounds,

0.04 metric ton;

= 80 pounds of coal.

3 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
why can light from the sun cause skin cancer but the light from a light bulb cannot cause skin cancer?
meriva

Answer:

Damage from UV exposure is cumulative and increases your skin cancer risk over time. While your body can repair some of the DNA damage in skin cells, it can't repair all of it. The unrepaired damage builds up over time and triggers mutations that cause skin cells to multiply rapidly. That can lead to malignant tumors.

4 0
2 years ago
Read 2 more answers
The acceleration due to gravity on the Moon's surface is
Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

g'=\frac{g}{6}

We know that weight of an object on Earth is,

w = m\times g

m = \frac{w}{g}

Similarly, weight on Moon will be

w' = m\times g'

w' = \frac{w}{g}\times\frac{g}{6}

w' = \frac{300}{6}

w' = 50

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0
3 years ago
We can model a lightning bolt as a very long, straight wire. If a lightning bolt carries a current of 30 kA, and you are unfortu
Liula [17]

Answer:

Magnetic field experienced = 4.5 × 10⁻⁴ T

Explanation:

The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by

B = (μ₀I)/(2πr)

B = ?

I = 20 KA = 20000 A

r = 8.9 m

μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A

B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T

8 0
3 years ago
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