Ask question

Ask question

All categories

3 years ago

6

A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e

nergy. The mass of an electron is 9.109 × 10−31 kg. Answer in units of m/s.

2 answers:

aleksandrvk [35]3 years ago

8 0

Answer:

4.66 × 10^5 m/s.

Explanation:

Given:

Mass of the electron, m = 9.109 × 10^−31 kg

KE = 0.991 × 10^−19 J.

Kinetic energy, KE = 1/2 × m × v^2

v = sqrt((2 × KE)/m)

= sqrt(2.176 × 10^11)

= 4.66 × 10^5 m/s.

alisha [4.7K]3 years ago

4 0

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

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

K.E= kinetic energy = $0.991\times 10^{-19}J$

m= mass of an electron = $9.109\times 10^{-31}kg$

v= velocity of object = ?

Putting in the values in the equation:

$0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}$

$v=466462m/s$

The speed necessary for the electron to have this energy is 466462 m/s

You might be interested in

If you had only one match, and entered a dark room containing an oil lamp, some newspaper, and some kindling wood, which would y

alukav5142 [94]

The match
you need to light the match before you can light anything else.

and after the match is lit, maybe light the oil lamp first

7 0

3 years ago

Read 2 more answers

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

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

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago

The unit used to measure electric current is the ampere (A). Now, assume that the current delivered at a wall socket reaches the

pav-90 [236]

Answer:

T = 0.017s

Explanation:

period is the time it takes a particle to make one oscillation

An electric current is periodic in nature

The current reaches 3.8A ten times.

So there must have been 10 cycles (10 periods) in 0.17s. let 'T' be the period:

$T=\frac{t}{n}$

t is the total time interval

n is the number of oscillations

$T=\frac{0.17}{10}$

10T = 0.17

T = 0.17/10 = 0.017s

8 0

3 years ago

Please help me thank you !!!!

Fantom [35]

Answer:

<em>UP</em>

Explanation:

heat flows from higher level to lower level

( higher concentration to lower concentration )

and since temperature in above block is less than the lower block, the heat will flow from lower block to higher block .

( Up )

5 0

3 years ago

You are attempting to row across a stream in your rowboat. Your paddling speed relative to still water is 3.0 m/s (i.e., if you

Nataliya [291]

Answer:

Please check the attached file for the diagram

Explanation:

The velocity of the of the rowboat $V_{tot}$ through the river is the resultant velocity. It is obtained taking a vector sum of the velocity in still water and the velocity of the river.

There are several ways to take this vector sum, but the question makes it simple for us to use Pythagoras's theorem because the East and North directions are perpendicular to each other.

Hence;

$V_{tot}^2=V_{still}^2+V_{w}^2\\V_{tot}^2=3^2+4^2$

$V_{tot}=\sqrt{3^2+4^2}\\ V_{tot}=\sqrt{25}=5m/s$

6 0

3 years ago