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Licemer1 [7]
3 years ago
14

Struggling with this. Can you help?

Physics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer:

Unclear without more information

Explanation:

This is about transfer of energy. If you assume there is no energy lost to the environment (virtually impossible) - then all of the potential energy at the beginning would be transferred to kinetic energy along the route.

So, assuming that the rollercoaster has a velocity of 0 m/s at point A, then you can calculate the potential energy using: Ep =mgh_A

At the point D, you would assume that the energy is split between some kinetic energy and some potential energy. So, we could say that the total energy is the sum of these: Etot=Ep+Ek=mgh_{D} +\frac{mv^2}{2}

If we assume all energy is transferred (ie. no energy lost to friction/heat etc) then we can equate these two and solve:

mgh_A=mgh_{D} +\frac{mv^2}{2}\\\\h_D=\frac{(2gh_A-v^2)}{2g} =22.4

However, this answer seems unlikely given the drawing - which implies that there is perhaps more information that is missing??

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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
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Each compression on a longitudinal wave correspond to what point on a transverse wave
krek1111 [17]

Explanation:

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In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur
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Answer:

a) C = (5-4D)/3

Explanation:

Given the simultaneous equation

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Using equation 1:

3C+4D = 5

Make C the subject of the formula:

Subtract 4D from both sides of the equation.

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3C = 5-4D

Divide both sides by 3:

3C/3 = (5-4D)/3

C = (5-4D)/3

Hence the expression that would result from the process is C = (5-4D)/3

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