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Serhud [2]
3 years ago
6

Two girls are standing on a rollerskates holding a basketball. One girl throws the ball to the other. Explain what happens to th

e girls with references to newton's third law of motion
Physics
1 answer:
katovenus [111]3 years ago
6 0

Answer:

Pretty sure that the they both go backwards because of the basketballs motion

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A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of
Anna007 [38]

Answer:

The ratio is  \frac{F_{T1}}{F_{T2}}  =  2

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

      Considering the second car

 The force causing it s movement  is mathematically represented as

       F_{T2} =  ma

 Considering the first car

 The force causing it s movement  is mathematically represented as

      F  = F_{T1} -F_{T2} = ma

=>   F_{T1} -ma  = ma

=>   F_{T1} =  2 ma

=> \frac{F_{T1}}{ma}  =  2

=> \frac{F_{T1}}{F_{T2}}  =  2

7 0
3 years ago
Please Help me with any of these possible.
Fiesta28 [93]

alll factor is possible

4 0
3 years ago
Which changes in an electric motor will make the motor stronger? Check all that apply. using a stronger permanent magnet using a
miv72 [106K]
Use stronger magnets
increase current
push magnets closer to coil
adding more sets of coils
4 0
3 years ago
Read 2 more answers
Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp
UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

6 0
3 years ago
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l
butalik [34]

Answer:

a)   f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} } , b)   Δf = 2 f₀ \frac{v}{c}

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo\frac{c+v}{c}

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ \frac{c}{ c-v}

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   \frac{c}{c-v}

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} }

         

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 ( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}

 

                f ’’ = fo ( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )

                f ’’ = fo ( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )

leave the linear term

               f ’’ = f₀ + f₀ 2\frac{v}{c}

the sound difference

               f ’’ -f₀ = 2f₀ v/c

               Δf = 2 f₀ \frac{v}{c}

4 0
3 years ago
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