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2 years ago

What relation does distance have with force and mass

1 answer:

4 0

This relationship was first published by Sir Issac Newton. His law of universal gravitation says that the force (F) of gravitational attraction between two objects with Mass1 and Mass2 at distance D is: F = G(mass1*mass2)/D squared

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Identify the two pieces of information you need to know the velocity of an object

Georgia [21]

Manuel with and height?

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2 years ago

What is the most widely accepted model used to predict the future of the universe?

lukranit [14]

Answer A an Open Universe

The future of universe is predicted from space time curve. There are many possibilities solutions or outcomes which on plotting are classified as open, close, flat and curved. Open Universe is when universe is expanding and it will continue to expand forever. This is proved experimentally as well that universe is expanding using red and blue shift observations from various aspects.

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2 years ago

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A petrol tanker ha 2800kg when empty and hold 30m3 0f petrol when full. The denity of petrol i 740kg/m3. Calculate the ma of the

White raven [17]

The mass of the tanker with petrol is 250000 N.

We are given that,

Mass of tanker= m =2800 kg

Volume of petrol= v =30 m³

Density of petrol= d =740 kg/m³

Thus , mass , density and volume relation can be given as,

density= mass/ Volume

Mass = Density × Volume

Mass = 740× 30

Mass = 22200 kg

The mass of the petrol is 22200 kg.

Total mass of tanker with petrol = Mass of petrol + Mass of tanker

Total mass of tanker with petrol= 22200+ 2800 kg

Total mass of tanker with petrol= 25000 kg

Total weight of the tanker with petrol = Mass of tanker full of petrol× g

Where, weight = m × g ,(g =10m/s²)

Total weight of the tanker with petrol= 25000×10 = 250000 N

Therefore, the mass of petrol, total mass of tanker with petrol and weight of tanker with petrol would be 22200 kg, 25000 kg and 250000 N.

To know more about mass

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11 months ago

Why can't fossils form in igneous rocks Magma flows too quickly for it to settle in The extreme heat destroys all of the remains

AleksandrR [38]

Answer:

you are right

Explanation:

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2 years ago

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A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$