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3 years ago

10

expain why different atoms of the same element always have the same atomic number but can have different mass numbers what are t

hese different atoms called ?

1 answer:

anzhelika [568]3 years ago

4 0

The atomic number tells you the number of protons in an atom. This value never changes because the number of protons in the nucleus always remains constant. The mass number tells you the number of protons and neutrons (or nucleons) together: protons + neutrons = mass number. Since the number of neutrons in the nucleus varies, you can have different amounts of neutrons in the same type of atom. These varied types are called isotopes.

Hope this helps.

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How do you find the rest mass (kg) of a 3.1 eV electron?

Scilla [17]

Answer:

Explanation:

The rest energy of any substance is defined by the Einstein's mass energy equivalence relation. Thus the rest mass of a electron is 9.11x10^-31 kg. The speed of light is 299,792,458 m/s. Thus multiplying the square of speed of light with the rest mass of electron gives the rest energy of the electron.

3 0

3 years ago

A motorboat accelerates uniformly from a velocity of 6.5 m/s to the west to a velocity of 1.5 m/s to the west. If its accelerati

monitta

Answer:

The displacement of the boat is 7.41 m

Explanation:

Given;

initial velocity of the motorboat, u = 6.5 m/s west

final velocity of the motorboat, v = 1.5 m/s west

acceleration of the motorboat, a = -2.7 m/s² east

The displacement of the boat is given by;

v² = u² + 2ad

where;

d is the displacement of the motorboat

1.5² = 6.5² + 2(-2.7)d

1.5² - 6.5² = -5.4d

-40 = -5.4d

d = (40) / 5.4

d = 7.41 m

Therefore, the displacement of the boat is 7.41 m

6 0

4 years ago

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV

Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

The radius of the dish is $R = 18cm$

Explanation:

From the question we are told that

The radius of the orbit is = $R = 35,000km = 35,000 *10^3 m$

The power output of the power is $P = 1 kW = 1000W$

The electric vector amplitude is given as $E = 0.1 mV/m = 0.1 *10^{-3}V/m$

The area of thereciever is $A_R = 5cm^2$

Generally the intensity of the dish is mathematically represented as

$I = \frac{P}{A}$

Where A is the area orbit which is a sphere so this is obtained as

$A = 4 \pi r^2$

$= (4 * 3.142 * (35,000 *10^3)^2)$

$=1.5395*10^{16} m^2$

Then substituting into the equation for intensity

$I_s = \frac{1000}{1.5395*10^{16}}$

$= 6.5*10^ {-14}W/m2$

Now the intensity received by the dish can be mathematically evaluated as

$I_d = \frac{1}{2} * c \epsilon_o E_D ^2$

Where c is thesped of light with a constant value $c = 3.0*10^8 m/s$

$\epsilon_o$ is the permitivity of free space with a value $8.85*10^{-12} N/m$

$E_D$ is the electric filed on the dish

So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

Now making the eletric field intensity the subject of the formula

$E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }$

substituting values

$E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }$

$= 7*10^{-6} V/m$

The incident power on the dish is what is been reflected to the receiver

$P_D = P_R$

Where $P_D$ is the power incident on the dish which is mathematically represented as

$P_D = I_d A_d$

$= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)$

And $P_R$ is the power incident on the dish which is mathematically represented as

$P_R = I_R A_R$

$= \frac{1}{2} c \epsilon_o E_R^2 A_R$

Now equating the two

$\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R$

Making R the subject we have

$R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }$

Substituting values

$R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }$

$R = 18cm$

8 0

3 years ago

A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretc

Anvisha [2.4K]

Explanation:

Given that,

Length of the spring, l = 50 cm

Mass, m = 330 g = 0.33 kg

(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :

$kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m$

(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.

(C) The frequency of the oscillation is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz$

5 0

4 years ago

When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If

masya89 [10]

The concept required to solve this problem is quantization of charge.

First the number of electrons will be calculated and then the total mass of the charge.

With these data it will be possible to calculate the percentage of load in the mass.

$Q= ne$

Here Q is the charge, n is the number of electrons and e is the charge on the electron

$n = \frac{Q}{e}$

Replacing,

$n = \frac{4*10^{-6}C}{1.6*10^{-19}}$

$n = 2.5 * 10^{13}77$

According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron

The total mass of the charge is

$m= nm_e$

Here,

m = Mass of the charge

n = Number of electrons

$m_e$ = Mass of the electron

$\text{Percentage change} = \frac{nm_e}{M}*100$

Replacing we have

$\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100$

$\text{Percentage change} = 6.9*10^{-14} \%$

6 0

4 years ago