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Ratling [72]
3 years ago
13

7. Mac and Tosh are arguing about the track design. Mac claims that the car is moving fastest at point F because it is furthest

along the track Tosh disagrees, claiming that the car moves fastest at point F because point F is at the lowest height Who do you agree with? Make a modification of the track design for point F and gather some evidence to support one claim or the other. Then discuss what changes you made, what observations were made, and the reasoning that supports one of the claim of either Mac or Tosh.​
Physics
1 answer:
olganol [36]3 years ago
7 0

Answer:

  • Tosh is correct.
  • Swap the locations of points B and F on the track and gather speed data. If Mac is correct, the speed at F (closer to the start) should be lower. (It will not be, confirming Tosh's claim.)

Explanation:

The total energy of the car is continuously being exchanged between potential energy and kinetic energy as the car moves along the track. Neglecting energy loss due to friction, the kinetic energy will be greatest when the potential energy is least, at the lowest point on the track. As a consequence we agree with Tosh that the speed will be greatest at F because it is the lowest point.

__

If the track were modified to move the lowest point nearer the start, say by interchanging points B and F, then data could be gathered to show whose theory is supported. The evidence needed is the speed of the car at the new location of point F. Tosh's argument is supported if the speed at the new point F is substantially the same.

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1. Una bola de 0.510 kg de masa se mueve al este (dirección +x) con una rapidez de 4.80 m/s y choca frontalmente con una bola de
Grace [21]

Responder:

3,37 m / s, + ve x - dirección

Explicación:

Utilizando la ley de conservación de la cantidad de movimiento expresada por la fórmula;

m1u1 + m2u2 = (m1 + m2) v

m1 y m2 son las masas de los objetos

u1 y u2 son sus velocidades iniciales

v es su velocidad común

Dado

m1 = 0,519 kg

u1 = 4,80 m / s

m2 = 0,220 kg

u2 = 0 m / s (cuerpo en reposo)

Necesario

Velocidad común v

Sustituir en la fórmula los valores dados;

0,519 (4,8) + 0,22 (0) = (0,519 + 0,220) v

2,4912 + 0 = 0,739 v

2,4912 = 0,739v

Dividir ambos lados por 0,739

2,4912 / 0,739 = 0,739 / 0,739

<em>v = 3,37 m / s </em>

<em>Por lo tanto, la rapidez de ambas bolas después de la colisión es de 3.37 m.s hacia la dirección x positiva, ya que m1> m2 y la velocidad común es positiva.</em>

6 0
3 years ago
An 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching th
ddd [48]

Answer:

1626.4 N

Explanation:

Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?

The parameters to be considered are:

Distance S = 3m

Time t = 0.55s

Since the man started from rest, initial velocity u = 0

Using second equation of motion

S = Ut + 1/2at^2

3 = 1/2 × a × 0.55^2

3 = 1/2 × a × 0.3025

a = 3/ 0.15125

a = 19.83 m/s^2

Force = mass × acceleration

Force = 82 × 19.83

Force = 1626.4 N

Therefore, the force that water exerted on him is 1626.4 N

4 0
3 years ago
If you increase the frequency of a sound wave four times what will happen to its speed
Anika [276]

Answer: The correct answer is "the speed of the wave becomes four times".

Explanation:

The relation between the speed, frequency and the wavelength is as follows:

v=f\lambda

Here, v is the speed of the wave, f is the frequency and \lambda is the wavelength.

The speed of the sound wave is directly proportional to the frequency.

In the given problem, if the speed of the sound wave is increased four times then the speed of the sound becomes four times.

Therefore, the speed of the sound wave becomes four times.

8 0
3 years ago
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Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t
Step2247 [10]

Answer:

\Delta p = 90.7 kPa

Explanation:

specific gravity of oil is = \frac{\rho_{oil}}{\rho_w}

\rho_{oil} = 0.85*1000 = 850 kg/m3

we know that

change in pressure  for oil is given as

\Delta p = \rho gh

here density and h is for oil

\Delta p = 850*5 *9.81 = 41,692.5 kPa

change in pressure  for WATER is given as

\Delta p = \rho gh

here density is for water and h is for water

\Delta p = 1000*5 *9.81 = 49,050 kPa

pressure change due to both is given as

\Delta p = 41692.3 + 49050 = 90742.5 N/m2

\Delta p = 90.7 kPa

8 0
3 years ago
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A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
Kisachek [45]

Answer:

The force is  F= 46.25kN

Explanation:

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

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

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                      v_2 = \frac{R}{A_n}

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                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

\rho is he density of water with value \rho =1000 kg /m^3

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

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At Equilibrium the summation of the of force on the horizontal axis is zero

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=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

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 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

=>                        F_x = 40.69kN

   Hence the force acting on the flange bolts required to hold the nozzle in place is

                      F = \sqrt{F_x^2 + F_y^2}

                         = \sqrt{40.69 ^2 + 21.99^2}

                         F= 46.25kN

                 

6 0
3 years ago
Read 2 more answers
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