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4 years ago

1) An object is hung using a metal spring. If now a current is passed through the spring, what will happen to this system?

Physics

1 answer:

son4ous [18]4 years ago

6 0

Answer:

D) The spring will contract, raising the weight.

Explanation:

According to the statement there is current that will enter the current through the metal ions that it has in its stratum. The passage of the current will generate within the spring a magnetic field that travels in a loop. That is, while the upper part of the spring which is also that of the spring acts as a north pole, the lower part of the spring and the magnetic field will act as the south pole. The position of the poles will generate an opposition effect that will generate an attraction to each other which will generate a contraction in the spring and an increase in weight on it.

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Can any1 tell if my answer is right

jok3333 [9.3K]

Answer:

The correct answer would be C. 5.0kg

Explanation:

The mass of an object never changes unless parts of the object are taken away. In other words, although the gravitational force is different on the moon then on the earth the mass of the object would remain the same.

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2 years ago

Why does an archaeologists use master tree ring patterns?

Alchen [17]

Answer:

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3 years ago

What are lowland areas on the moon

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3 years ago

The voltage applied across a given parallel-plate capacitor is doubled. How is the energy stored in the capacitor affected?

ikadub [295]

Answer:

The energy stored in the capacitor quadruples its original value.

Explanation:

The energy stored in a capacitor is given by the equation

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V is the voltage across the plates

The capacitance, C, depends only on the properties of the capacitor, so it does not change when the voltage applied is changed.

Instead, in this problem the voltage applied is doubled:

V' = 2V

So the new energy stored is

$U'=\frac{1}{2}C(2V)^2=4(\frac{1}{2}CV^2)=4U$

so, the energy stored has quadrupled.

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3 years ago

A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, an

Oxana [17]

Answer:

Explanation:

Mass of 190kg

Coefficient of static friction is 0.4

Coefficient of kinetic friction 0.36

Horizontal force= 500N

Taking g=9.81m/s^2.

The weight of the body my

W=190×9.81=1863.91N

There is a normal acting on the body which is equal to the weight

N=W=1863.91N

Frictional force(fr) is acting on the body and it is opposite the horizontal force.

The minimum force to be overcome before the object can start to move is Fr = μsN

Fr= μsN. μs=0.4

Fr= 0.4×1863.91

Fr=745.56N.

Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.

Then the frictional force at that time is equal to the horizontal force

Therefore

Functional force = 500N

b. Mass of asteroid is

M=2000kg

Asteroid velocity at a particular instant is,

U=(-1.30x10^4, 4.20x10^4, 0)m/s

Magnitude of U is

U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0

U=√1.933E9

U=4.39×10^4m/s

Position of the asteroid from the centre of the earth is,

R= (6.00x10^6, 10.00x10^6, 0)m.

The magnitude of the radius is

R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2

R=√3.6E13+10E13+0

R=√13.6E13

R=1.17E7m

R^2=13.6E13m

The mass of the earth is

Me=5.97x10^24 kg

The momentum of the asteroid after time, t=1.5×10^3s

Given that G=6.67x10^-11Nm^2/kg^2

Momentum is

Mv-Mu=Ft

There the new momentum will be

Mv=Ft+Mu

Now we the to find the force the earth exert on the asteroid by using

F=GMMe/R^2

F=6.67E-11 ×2000× 5.97E24 /13.6E13

F=7.964E17/13.6E13

F=5855.88N

The new momentum

Mv= Mu+Ft

Mv= 2000(4.39E4)+5855.88(1.5E3)

Mv=9.66E7kgm/s

The new momentum is 9.66×10^7 Kgm/s

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3 years ago