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Crank
3 years ago
12

What's the answer to 28-4(x-5)=-4

Physics
2 answers:
IceJOKER [234]3 years ago
7 0
28-4(x-5)=-4
28-4x-20=-4
+28         +28
-------------------
-4x-20=24
      +20 +20
-------------------
-4x=44
-4  -4
---------
x=-11
Anna11 [10]3 years ago
6 0
The answer is -23.

I hope this will be of use to you
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Determine the weight of an average physical science textbook whose mass is 3.1 kilograms. The acceleration due to gravity is 9.8
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Weight equals mass*gravity
W = mg

Given m = 3.1 kg, g = 9.8 m/s^2

W = (3.1)(9.8)
W = 30.38
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Determine the energy, in electron volts, of an emitted photon when an electron transition from n=3 to n=2 occurs
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. What is the single most important equation in all of physics?​
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Answer: F = ma,

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Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed
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Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.

As, we know that the same charges repeal each other both of the balloons with be apart from each other.

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3 0
3 years ago
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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
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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
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