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3 years ago

For the vectors in the figure, with a = 16, b = 12, and c = 20 what are (a) the magnitude and (b) the direction of

1 answer:

5 0

Probably need to see the figure, although a=4x4, b = 3x4, c=5x4 which suggests to me that the figure might be a 3,4,5 right angled triangle ?

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larisa [96]

Answer:

wow i have know idea

Explanation:

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3 years ago

The switch on the electromagnet, initially open, is closed. What is the direction of the induced current in the wire loop (as se

Marianna [84]

Answer:

The induced current is clockwise

6 0

3 years ago

PLSS HELP!! Are these True or False questions

skelet666 [1.2K]

Answer:

11. True

12. False (air pressure)

13. False (nitrogen)

14. False (stratosphere)

Explanation:

5 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

Consider an electrical transformer has 10 loops on its primary coil and 20 loops on its secondary coil. What is the voltage in t

Arlecino [84]

Answer:

240 V

Explanation:

number of turns in primary coil, Np = 10

Number of loops in secondary coil, Ns = 20

Voltage in primary coil, Vp = 120 V

Let the voltage in secondary coil is Vs.

So, Vs / Vp = Ns / Np

Vs / 120 = 20 / 10

Vs / 120 = 2

Vs = 240 V

Thus, the voltage in secondary coil is 240 Volt.

4 0

3 years ago