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vova2212 [387]
4 years ago
8

An action force is equal in _____ and opposite in _____ to a reaction force. A. power; direction B. direction; power C. directio

n; magnitude D. magnitude; direction
Physics
2 answers:
butalik [34]4 years ago
4 0

The correct answer is D. magnitude; direction

We know this thanks to Newton's 3rd law

Hope that helps!!

tatiyna4 years ago
4 0

Magnetuide; direction hope that helps

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A machine has an efficiency of 80%. How much work must be done on the machine so to make it do 50,000 J of output work?
Tamiku [17]

Work formula is Work=N∙M or Joule (J)

So you have the following given:

50 000 is the output work

.8 is the efficiency

 if you input the given = 50000 = .8*J

To get the answer just divide 50000 by .8 to get the answer.

62,500 J is the amount of work to be done.

8 0
3 years ago
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Find the momentum of a 3.0 kg mass when it is stopped
lina2011 [118]

Answer:

4,0kg it is stopped!!!!!

3 0
3 years ago
A transverse traveling wave on a cord is represented by, where D and x are in meters and t is in seconds.
yarga [219]

Answer with Explanation:

We are given that a transverse travelling wave on a cord is represented by

D=0.51sin(6.1x+76t)

Where D and x in meters

t(in seconds)

a.General equation of transverse wave

y=Asin(kx+\omega t)

By comparing we get

A=0.51

k=6.1

k=\frac{2\pi}{\lambda}

Wavelength,\lambda=\frac{2\pi}{k}=\frac{2\pi}{6.1}=1.03

b.\omega=76

Frequency,f=\frac{\omega}{2\pi}=\frac{76}{2\pi}=12.096Hz

c.Velocity,v=f\lambda=12.096\times 1.03=12.46m/s

Direction:Towards negative x- axis

d.Amplitude,A=0.51 m

e.Maximum speed,v_{max}=A\omega=0.51\times 76=38.76 m/s

Minimum speed,v_{min}=0

6 0
4 years ago
Which is an example of radiation?
Y_Kistochka [10]
A common example of radiation would have to be sunlight

6 0
3 years ago
Read 2 more answers
Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out
PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is  

\lambda _2 = (6.48,6.52) \ cm

Explanation:

From the question we are told that

     The wavelength of the note is \lambda  =  6.50 \ cm = 0.065 \ m

     The difference in beat frequency is \Delta  f = 17.0 \ Hz

     

Generally the frequency of the note played by the guitar that is in tune is  

        f_1 = \frac{v_s}{\lambda}

Where v_s is the speed of sound with a constant value v_s  =  343 \ m/s

       f_1 = \frac{343}{0.0065}

      f_1 = 5276.9 \ Hz

The difference in beat is mathematically represented as

       \Delta  f =  |f_1 - f_2|

Where f_2 is the frequency of the sound from the out of tune guitar

     f_2 =f_1  \pm \Delta f

substituting values

      f_2 =f_1 + \Delta f

      f_2 = 5276.9 + 17.0  

     f_2 = 5293.9 \ Hz

The wavelength for this frequency is

      \lambda_2 = \frac{343 }{5293.9}

     \lambda_2 = 0.0648 \ m

    \lambda_2 = 6.48 \ cm

For the second value of the second frequency

     f_2 =  f_1 - \Delta f

     f_2 = 5276.9 -17

      f_2 = 5259.9 Hz

The wavelength for this frequency is

   \lambda _2 = \frac{343}{5259.9}

   \lambda _2 = 0.0652 \ m

   \lambda _2 = 6.52 \ cm

8 0
3 years ago
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