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Tanya [424]
3 years ago
12

A mass of 10 kg is at a point A on the table. It moved to a point B. If the line joining A and B is horizontal. What is the work

done on the object by the gravitational force ? Explain your answer
Physics
1 answer:
elena55 [62]3 years ago
6 0

Answer:

Work done by gravitational force = 0 joule

Explanation:

Work done is given by the relation

                        W  =  F   x    S   joule

Where,    F  -  the force applied in the form of push or pull

               S  -  displacement caused by the force

If a force is acting on a body and it doesn't cause any displacement, then work done will be zero.

Gravitational force acts on the body even if the body is at rest.

<em>Work done by the gravitational is applicable only if there is some vertical component of motion involved.</em>

In this case, some amount of work is done to move the body from point A to B.But, the body is displaced in the horizontal direction. No vertical motion is involved.

So, the work done due to gravitational force on a mass is zero joule.

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Ferromagnetic materials-
QveST [7]

Answer:

materials which exhibit a spontaneous net magnetization at the atomic level, even in the absence of an external magnetic field.

Explanation:

When a material is placed within a magnetic field, the magnetic forces of the material's electrons will be affected. This effect is known as Faraday's Law of Magnetic Induction. However, materials can react quite differently to the presence of an external magnetic field. This reaction is dependent on a number of factors, such as the atomic and molecular structure of the material, and the net magnetic field associated with the atoms. The magnetic moments associated with atoms have three origins. These are the electron motion, the change in motion caused by an external magnetic field, and the spin of the electrons.

5 0
3 years ago
The combined-gas law relates which of these?
Fofino [41]
The combined-gas law relates which temperature, pressure and volume.

Temperature=T
Pressure=P
Volume=V

(P₁*V₁) / T₁=(P₂*V₂) / T₂

D. Temperature, pressuere and volume.
5 0
3 years ago
6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t
spin [16.1K]

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.    

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.            

5 0
3 years ago
Read 2 more answers
You decide to ride your bike for 30 minutes, raising your heart rate to the middle of your target heart rate zone. What type of
a_sh-v [17]
It would have to be cardio i think not totally sure
4 0
4 years ago
Read 2 more answers
Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

P_{2} > P_{1}.

6 0
4 years ago
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