An aircraft travels a distance of 50km in a straight line

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is s

kumpel [21]

To solve this problem we must keep in mind the concepts related to angular kinematic equations. For which the angular velocity is defined as

$\omega_f =\omega_i-\alpha t$

Where

$\omega_f =$ Final angular velocity

$\omega_i =$ Initial angular velocity

$\alpha =$ Angular acceleration

t= time

In this case we do not have a final angular velocity, then

$\omega_i = \alpha t$

Re-arrange for $\alpha$

$\alpha= \frac{\omega_i}{t}$

$\alpha = \frac{910}{0.167}$

$\alpha = 5449.1 rad\s^2$

Therefore the mangitude of the angular aceleration is 5449.1rad/s²

6 0

4 years ago

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect

klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = \frac{1}{2}mv^2$

Making v the subject

$v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $\frac{mv^2}{r}$ ] and this is mathematically represented as

$Bqv = \frac{mv^2}{r}$

making B the subject

$B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

$B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

$= 0.0048 T$

7 0

4 years ago

A rocket is launched upward with a constant acceleration of 165 m/s^2. After 8.00 seconds of ascension a passenger on the rocket

Novay_Z [31]

To solve this problem we will apply the concepts related to the linear kinematic movement. We will start by finding the speed of the body from time and the acceleration given.

Through the position equations we will calculate the distance traveled.

Finally, using this same position relationship and considering the previously found speed, we can determine the time to reach your goal.

For time (t) and acceleration (a) we have to,

$t = 8s, a = 165m/s^2$