Ask question

Ask question

All categories

Free_Kalibri [48]

3 years ago

7

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate

the charge on the ion.

1 answer:

riadik2000 [5.3K]3 years ago

4 0

Answer: $6.408(10)^{-19} C$

Explanation:

This problem can be solved by the following equation:

$\Delta K=q V$

Where:

$\Delta K=7.37(10)^{-17} J$ is the change in kinetic energy

$V=115 V$ is the electric potential difference

$q$ is the electric charge

Finding $q$ :

$q=\frac{\Delta K}{V}$

$q=\frac{7.37(10)^{-17} J}{115 V}$

Finally:

$q=6.408(10)^{-19} C$

You might be interested in

Is motion on earth different from space

Rus_ich [418]

It's not that the laws of motion are any different on Earth than in space. But Earth's gravitational field has such an overwhelming force it masks their precise effects. And gravity is integral to all sorts of phenomena that we take for granted.

8 0

2 years ago

A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

Alexandra [31]

Answer:

The maximum no. of electrons- $2.25\times 10^{22}$

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

$I = \frac{Q}{t}$

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

$Q = 1.0\times 3600 = 3600\ C$

Maximum number of electrons, n is given by:

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

where

e = charge on an electron = $1.6\times 10^{- 19}\ C$

Thus

$n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}$

3 0

3 years ago

Why are solids good at transferring heat through conduction?

seraphim [82]

The chemical bonds alow it to pass to the other object.

6 0

3 years ago

Read 2 more answers

A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a

Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = $2\pi$ radians

So, 2.73 revolutions = $2.73\times 2\pi=17.153\ radians$

Therefore, the angular velocity of the tack is, $\omega=17.153\ rad/s$

Now, radial acceleration of the tack is given as:

$a_r=\omega^2 r$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]$

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0

3 years ago

A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele

ludmilkaskok [199]

Answer:

4. Downward and its value is constant

Explanation:

As this is a case of projectile motion, we use the reference frame where upward direction to be positive for $y$ , and in the same way to be negative in the downward direction. On another hand, we have that gravity is always acting this means that gravitational acceleration g is directed downward constantly over the dart not only during the upward but also during the downward part of the trajectory. And it is ruled by the following equations.

For the x-axis

$v_{x}=v_{0}cos(45\°)=constant$

$x=(v_{0}cos(45\°))t$

For the y-axis

$v_{y}=v_{0}sin(45\°)-gt$

$y=v_{0}sin(45\°)t-\frac{1}{2} gt^{2}$

Where $v_{0}$ , is the initial velocity.

8 0

3 years ago