How fast is a 5 kg object traveling that has 20 kg m/s of momentum?

A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni

irinina [24]

The magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

The given parameters;

length of the solenoid, L = 91 cm = 0.91 m
radius of the solenoid, r = 1.5 cm = 0.015 m
number of turns of the solenoid, N = 1300
current in the solenoid, I = 3.6 A

The magnitude of the magnetic field inside the solenoid is calculated as;

$B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\$

where;

$\mu_o$ is the permeability of frees space = 4π x 10⁻⁷ T.m/A

$B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T$

Thus, the magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

Learn more here:brainly.com/question/17137684

7 0

2 years ago

Https://phet.colorado.edu/sims/html/balloons-and-static-electricity/latest/balloons-and-static-electricity_en.html

Katarina [22]

Answer:

there is not pic

Explanation:

5 0

2 years ago

Work is being done in which of these situations? all motions are at a constant velocity

fredd [130]

Where the force is not perpendicular to the path of motion

are you missing the the situations ?

7 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

iVinArrow [24]

Answer:

D

C

Explanation: BRAINLIEST ME

6 0

3 years ago

An automobile whose speed is increasing at a rate of 0.800 m/s2 travels along a circular road of radius 10.0 m.

Ket [755]

(a) $a_t = 0.800 m/s^2$

The tangential acceleration component of the car is simply equal to the change of the tangential speed divided by the time taken:

$a_t = \frac{\Delta v}{\Delta t}$

This rate of change is already given by the problem, 0.800 m/s^2, so the tangential acceleration of the car is

$a_t = 0.800 m/s^2$

(b) $a_c = 0.9 m/s^2$

The centripetal acceleration component is given by

$a_c = \frac{v^2}{r}$

where

v is the tangential speed

r is the radius of the trajectory

When the speed is v = 3.00 m/s, the centripetal acceleration is (the radius is r = 10.0 m):

$a_c = \frac{(3.00 m/s)^2}{10.0 m}=0.9 m/s^2$

(c) $1.2 m/s^2, 48.4^{\circ}$

The centripetal acceleration and the tangential acceleration are perpendicular to each other, so the magnitude of the total acceleration can be found by using Pythagorean's theorem:

$a=\sqrt{a_t^2+a_c^2}=\sqrt{(0.8 m/s^2)^2+(0.9 m/s^2)^2}=1.2 m/s^2$

and the direction is given by:

$tan \theta =\frac{a_c}{a_t}=\frac{0.9 m/s^2}{0.8 m/s^2}=1.125\\\theta=tan^{-1}(1.125)=48.4^{\circ}$

where the angle is measured with respect to the direction of the tangential acceleration.

4 0

3 years ago