Which statement correctly compares sound and light waves?

What does the delta symbol represent in the equation?

Alex_Xolod [135]

Answer:

Whenever you see the delta symbol infront of another, it always represents change. In other words:

Δu = u_f - u_i

u_i means "initial velocity"
u_f means "final velocity"

And in general, ΔX means Xfinal - Xinitial

6 0

3 years ago

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A bag of potatoes has a mass of 2.5 kg. What is its approximate weight on Earth?

ad-work [718]

Weight = m (mass) * g (acceleration due to gravity)

g = 9.80 m/s^2
m = 2.5 kg = 2,500 g

Weight = 2,500 g * 9.80 m/s^2
Weight = 24,500 N

5 0

3 years ago

Which statement about real gases is true? A) Forces of attraction and repulsion exist between gas particles at close range. B) T

olga55 [171]

Answer: A) Forces of attraction and repulsion exist between gas particles at close range.

Explanation:

The <u>Ideal Gas equation</u> is:

$P.V=n.R.T$

Where:

$P$ is the pressure of the gas

$V$ is the volume of the gas

$n$ the number of moles of gas

$R$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them. In this sense, real gases can behave approximately to an ideal gas, under conditions of high temperature and low pressures.

However, at low temperatures or high pressures, real gases deviate significantly from ideal gas behavior. This is because at low temperatures molecules begin to move slower, allowing the repulsive and attractive forces among them to take effect. In fact, <u>the attraction forces are responsible for the condensation of the gas</u>. In addition, at high pressures the volume of molecules cannot be approximated to zero, hence the volume of these molecules is not negligible anymore.

7 0

3 years ago

A rock from Mars experiences a 263% increase in its weight once it arrives on Earth. What accounts for this increase in weight?

leva [86]

The mass of Mars is smaller than that the mass of Earth, therefore the acceleration of gravity on Mars is much less than that the acceleration of gravity of the Earth.

The weight of an object is directly proportional to the magnitude of the gravitational acceleration of the planet where it is.

In this case, the gravitational acceleration in Mars is 3.711 m/s² and that of the planet Earth is 9.807 m/s².

The acceleration of gravity on Earth is 2.63 times greater than that the acceleration of gravity on Mars. This explains that the same object weighs 263% more on earth

3 0

3 years ago

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$