Ask question

Ask question

All categories

3 years ago

11

An aluminum wire of length 1.0 meter has a resistance of 9.0x10^-3 ohm. If the wire were cut into two equal lengths, each length

would have a resistance of

2 answers:

murzikaleks [220]3 years ago

0 0

Answer:since resistance depends on length, then the resistance will be shared equally since it was cut equally therefore the resistance is

4.5*10^-3 ohm

Explanation:

Diano4ka-milaya [45]3 years ago

0 0

Answer : The each length would have a resistance of $4.5\times 10^{-3}\Omega$

Explanation :

Let the resistance of equal length of wire be, r

That mean,

$r+r=R$

where,

r = resistance of equal length of wire

R = total resistance of wire = $9.0\times 10^{-3}\Omega$

As we know that resistance is directly proportional to length.

Total resistance of aluminum wire = Sum of resistance of two equal piece of wires

$r+r=9.0\times 10^{-3}\Omega$

$2r=9.0\times 10^{-3}\Omega$

$r=\frac{9.0\times 10^{-3}\Omega}{2}$

$r=4.5\times 10^{-3}\Omega$

Therefore, the each length would have a resistance of $4.5\times 10^{-3}\Omega$

You might be interested in

(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

PilotLPTM [1.2K]

Answer:

Part a)

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

$v = 2.18 \times 10^5 m/s$

Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

$T = 8.2 \times 10^{15} s$

Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

5 0

3 years ago

Infrared observations of the orbits of stars close to the galactic center indicate a small object at the center with a mass of a

MariettaO [177]

Answer:

(4.31±0.38) million Solar masses.

Explanation:

The galactic center is the center of the milky way around which the galaxy rotates. It is most likely the location of a supermassive black hole which has a mass of (4.31±0.38) million Solar masses. The location is called Sagittarius A*.

As there is interstellar dust in our line of sight from the Earth infrared observations need to be taken.

5 0

2 years ago

Why do heavier objects fall at the same speed as lighter ones?

GaryK [48]

Speed of any freely falling object is always same. Provided, both are left to fall from the same height. If you perform this experiment in a perfect vacuum or near vacuum laboratory, both of them will reach ground with same velocity this is because there is no resistance to their motion. This is always true no matter where you go and perform this experiment.
It can be easily proved from conservation of mechanical energy. Why conserving energy? because there are no forces acting on the freely falling objects other than conservative force(mg).

5 0

3 years ago

List five examples from daily life in which you see periodic motion caused by a spring or cord.

dmitriy555 [2]

car suspension

guitar string

clips for hanging clothes

keyboards buttons

elevators

4 0

2 years ago

Read 2 more answers

Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

Assoli18 [71]

Answer:

Part a)

$\rho = 3\epsilon_0 k r^2$

Part b)

$Q = 4\pi \epsilon_0kR^5$

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

$E = kr^3 \hat r$

now electric flux through a spherical surface of radius r is given as

$\phi = E. A$

$\phi = kr^3(4\pi r^2)$

now by Guass law we know that

$E.A = \frac{q}{\epsilon_0}$

$q = 4\pi \epsilon_0kr^5$

now volume charge density is given as

$\rho = \frac{q}{\frac{4}{3}\pi r^3}$

$\rho = 3\epsilon_0 k r^2$

Part b)

Total charge inside the radius R is given as

$Q = 4\pi \epsilon_0kR^5$

7 0

2 years ago