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2 years ago

Please help me find the equations guys

Physics

1 answer:

Alexeev081 [22]2 years ago

4 0

The line at the bottom of the picture ... probably the first line on a list of choices .. is the correct equation.

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A sample of helium (He) occupies 8.0 liters at 1 atm and 20.0◦C. What pressure is necessary to change the volume to 1.0 liters a

nevsk [136]

Apply the combined gas law

PV/T = const.

P = pressure, V = volume, T = temperature, PV/T must stay constant.

Initial PVT values:

P = 1atm, V = 8.0L, T = 20.0°C = 293.15K

Final PVT values:

P = ?, V = 1.0L, T = 10.0°C = 283.15K

Set the PV/T expression for the initial and final PVT values equal to each other and solve for the final P:

1(8.0)/293.15 = P(1.0)/283.15

P = 7.7atm

7 0

3 years ago

How does the increasing mass effect the force of an object in motion?

irina [24]

Answer:

According to second law of motion,the acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

So simply, it can be affected due to increasing force as there is close relationship between momentum.

Explanation:

The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.

I hope it was helpful for you :)

7 0

1 year ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$