Ask question

Ask question

All categories

3 years ago

11

If lithium has an atomic number of 3 and a mass number of 7, the correct number of protons, neutrons, and electrons, respectivel

y, is:

1 answer:

Karo-lina-s [1.5K]3 years ago

6 0

3 protons
3 electrons
4 neutrons (7-3)

You might be interested in

Consider a sample of gas in a container on a comfortable spring day in chicago, il. the celsius temperature suddenly doubles, an

Vinil7 [7]

To solve this problem, we must first assume that the gas acts like an ideal gas so that we can use the ideal gas equation:

P V = n R T

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T is the absolute temperature

Assuming that the number of moles is constant, then we can write all the variables in the left side:

P V / T = k where k is a constant (n times R)

Equating two conditions or two states:

P1 V1 / T1 = P2 V2 / T2

We are given that V2 = 2 V1 therefore

P1 V1 T2 = P2 (2V1) T1

P1 T2 = 2 P2 T1

Additionally we are given that the temperature in Celsius is doubled, however in the formula we use the absolute temperature in Kelvin, therefore:

T1 (K) = T1 + 273.15

T2 (K) = 2T1 + 273.15

and P1 = 12 atm

Substituting:

<span>12 (2T1 + 273.15) = 2 P2 (T1 + 273.15)</span>

P2 = 6 (2T1 + 273.15) / (T1 + 273.15)

Assuming that a nice spring day in Chicago has a temperature of 15 Celsius, therefore:

P2 = 6 (2*15 + 273.15) / (15 + 273.15)

<span>P2 = 6.312 atm</span>

3 0

3 years ago

What happens to the bulb when the battery changes from 1.5V to 9v?

timofeeve [1]

Answer:

it gets brighter because the volta increases

4 0

2 years ago

A 720 g softball is traveling at 15.0 m/s when caught. If the force of the glove on the ball is 520 N, what is the time it takes

Sholpan [36]

Answer:

The time it takes the ball to stop is 0.021 s.

Explanation:

Given;

mass of the softball, m = 720 g = 0.72 kg

velocity of the ball, v = 15.0 m/s

applied force, F = 520 N

Apply Newton's second law of motion, to determine the time it takes the ball to stop;

$F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.72 \ \times \ 15}{520} \\\\t = 0.021 \ s \\$

Therefore, the time it takes the ball to stop is 0.021 s.

5 0

3 years ago

Please indicate how long each bar is in centimeters and millimeters.

Maru [420]

Answer:

1) 74 cm or 740 mm

2) 33 cm or 330 mm

3) 76 cm or 760 mm

4) 17 cm or 170 mm

brainliest !?!?

6 0

2 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago