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3 years ago

If the strength of the magnetic field at B is 20 units, the strength of the magnetic field at A is _____.

Physics

2 answers:

xz_007 [3.2K]3 years ago

8 0

The answer is A = 20 units

Fantom [35]3 years ago

3 0

Answer: The answer is actually 80 units

Explanation: I did the assignment and 80 units was the answer that was correct.

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A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s.

Serggg [28]

Answer:

(a) α = 35.20 rad/s^2

(b) θ = 802°

(d) a = 156.64 cm/s^2

Explanation:

(a) To find the angular acceleration of the disc you use the following formula:

$\alpha=\frac{\omega-\omega_o}{t}$ (1)

w: angular speed of the disc = 31.4 rad/s

wo: initial angular speed = 0 rad/s

t: time = 0.892s

You replace the values of the parameters in the equation (1):

$\alpha=\frac{31.4rad/s-0rad/s}{0.892s}=35.20\frac{rad}{s^2}$

The angular acceleration of the disc, for the given time, is 35.20rad/s^2

(b) To calculate the angle describe by the disc in such a time you use:

$\theta=\frac{1}{2}\alpha t^2$ (2)

$\theta=\frac{1}{2}(35.20rad/s^2)(0.892s)^2=14.00rad$

In degrees you have:

$\theta=14.00rad*\frac{180\°}{\pi \ rad}=802\°$

The angle described by the disc is 802°

(c) To calculate the tangential speed of the microbe for t=0.892s, you use the following formula:

$v=\omega r$ (3)

w: angular speed for t = 0.892s = 31.4rad/s

r: radius of the disc = 4.45cm

$v=(31.4rad/s)(4.45cm)=139.73\frac{cm}{s}$

The tangential speed is 139.73 cm/s

(d) The tangential acceleration is calculated by using the following formula:

$a=\alpha r$

α: angular acceleration for t=0.892s

$a=(35.20rad/s^2)(4.45cm)=156.64\frac{cm}{s^2}$

The tangential acceleration is 156.64cm/s^2

7 0

2 years ago

The angle of incidence of a ray of light striking an equilateral triangular prisms ABC of refracting angle 60o is 40o. Calculate

valkas [14]

Answer:

1: the refracted angle in the first face is equal to the incident angle that is 60degrees

2. Emergence Angle is 42degrees

Explanation:

Pls see attached file

5 0

3 years ago

A 26.5-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 12.5-g object moving in the same dire

Dahasolnce [82]

Answer:

v₁ =0.19 m/s and v₂ = 0.18 m/s

Explanation:

By conservation of energy and conservation of momentum we can find the velocity of each object after the collision:

Momentum:

Before (b) = After (a)

$m_{1}v_{1b} + m_{2}v_{2b} = m_{1}v_{1a} + m_{2}v_{2a}$

$26.5 kg*0.205 m/s + 12.5 kg*0.150 m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}$ $7.31 kg*m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}$ (1)

Energy:

Before (b) = After (a)

$\frac{1}{2}m_{1}v_{1b}^{2} + \frac{1}{2}m_{2}v_{2b}^{2} = \frac{1}{2}m_{1}v_{1a}^{2} + \frac{1}{2}m_{2}v_{2a}^{2}$

$26.5 kg*(0.205 m/s)^{2} + 12.5 kg*(0.150 m/s)^{2} = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$ (2)

From equation (1) we have:

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg}$ (3)

Now, by entering equation (3) into (2) we have:

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*(\frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg})^{2}$ (4)

By solving equation (4) for $v_{1a}$ , we will have two values for

$v_{1a} = 0.16$

$v_{1a} = 0.21$

We will take the average of both values:

$v_{1a} = 0.19 m/s$

Now, by introducing this value into equation (3) we can find $v_{2a}$ :

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*0.19 m/s}{12.5 kg}$

$v_{2a} = 0.18 m/s$

Therefore, the velocity of object 1 and object 2 after the collision is 0.19 m/s and 0.18 m/s, respectively.

I hope it helps you!

6 0

2 years ago

In which beaker were the particles moving the most slowly?

viva [34]

Answer:

It might be wrong but d? i think

4 0

3 years ago

how does the index of refraction of an unknown substance compare to the index of refraction of a vacuum? explain how the speed a

Arturiano [62]

Answer:

By measuring the speed of light in both in space and that unknown substance

Explanation:

Refractive Index (Index of Refraction) is a value calculated from the ratio of the speed of light in a vacuum to that in a second medium of greater density.

By measuring the speed of light in that medium and dividing it by speed of light 3*10^8 we can measure the refractive index of any medium

8 0

2 years ago