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Juli2301 [7.4K]
3 years ago
15

What would decrease the resistance of wires carrying an electric current

Physics
2 answers:
Lostsunrise [7]3 years ago
6 0

Answer:

A. shorter wires

Explanation:

Murljashka [212]3 years ago
3 0
I believe the answer is A. shorter wires


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A spring with a spring constant of 110n/m is stretched by 12 cm in the positive direction. How much force is applied to the spri
taurus [48]

The force applied on the spring to stretch it is 13.2 N.

Hooke's law is a law of elasticity discovered by the English scientist Robert Hooke in 1660 that states that the displacement or size of a deformation is directly proportional to the deforming force or load for relatively small deformations of an object. When the load is removed under these conditions, the object returns to its original shape and size.

According to Hooke's law, F = k*e

where F is the force on the spring

k is force constant

and e is extension

F = (110)*(0.12)

F = 13.2 N

For more information on Hooke's law, visit :

brainly.com/question/13348278

#SPJ4

7 0
1 year ago
The diameter of Saturn is nearly ten times that of Earth. However, the density of Saturn is much less than that of Earth. What i
Tatiana [17]

Answer:

Saturn is a gaseous planet

7 0
3 years ago
When u write on a piece of glass sheet with a piece of chalk , the writing is not clear explain .
svetlana [45]
Becuse your weighting with chalk that has pigment
6 0
4 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
andreev551 [17]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0
3 years ago
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