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luda_lava [24]
3 years ago
11

As a consultant to the soft-drink industry, Dr. J is given the task of conducting the ultimate Pepsi taste test. This is Dr. J's

tenth taste test, which puts him seven up on his nearest consultant, who had only done three. Of course Dr. J is very qualified, having been hooked on soft drinks (especially orange soda) since he was Nehi to a pop bottle. Dr. J mounts a rather large container of Pepsi on a ledge some 3 meters above the ground. A bullet of mass 5 grams is then fired into the container, thus killing the taste. Not only that, but the Pepsi falls through the bullet hole onto the ground below (causing the taste to go flat). The wall of the container is 2 cm thick. The velocity of the bullet changes from an initial value of 500 m/sec just before striking the container wall to 5 m/sec upon leaving the container wall and entering the Pepsi. It finally fizzles out at a point 25 cm from the container wall.A. How much work does the container wall do on the bullet?
B. How much work does the Pepsi do on the bullet?
C. At what velocity does the Pepsi hit the floor?
Physics
1 answer:
dem82 [27]3 years ago
3 0

Answer:

A) How much work does the Pepsi do on the bullet = 0.0625J

B) At what velocity does the Pepsi hit the floor = 7.67m/s

Explanation:

  • Given mass of bullet = 5g
  • initial velocity = 500 m/sec
  • final velocity = 5 m/sec

From work done = Force X Distance

  • WD = Mad
  • a = v - u /t

Workdone ; m(v - u )/t

A) The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi:

  • but The work done by the container wall on the bullet is equal to the change in kinetic energy on either side of the wall: W = change in kinetic energy = 1/2 mv²

= 1/2 X 0.005 x 500² - 1/2 X 0.055 x 5²

= 624.94J

  • Therefore ; The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi = W = change in KE = 0.0625 - 0 = 0.0625J

B) At what velocity does the Pepsi hit the floor?

  • From conservation of energy principle; PE = KE
  • mgh = 1/2mv²
  • 0.98 X 3 = 1/2v²
  • velocity = 7.67m/s

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Answer: 361° C

Explanation:

Given

Initial pressure of the gas, P1 = 294 kPa

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Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K

Final temperature of the gas, T2 = ?

Let us assume that the gas is an ideal gas, then we use the equation below to solve

T2/T1 = P2/P1

T2 = T1 * (P2/P1)

T2 = (100 + 273) * (500 / 294)

T2 = 373 * (500 / 294)

T2 = 373 * 1.7

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3 years ago
What's the frequency of a wave with a wavelength of 10 and velocity of 200m/s?
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Answer:

\boxed {\boxed {\sf 20 \ Hz}}

Explanation:

The frequency of a wave can be found using the following formula.

f=\frac{v}{\lambda}

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The wavelength is 10 meters and the velocity is 200 meters per second.

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Therefore:

v= 200 \ m*s^{-1} \\\lambda = 10 \ m

Substitute the values into the formula.

f=\frac{200 \ m*s^{-1}}{10 \ m}

Divide and note that the meters (m) will cancel each other out.

f=\frac{200 \ s^{-1}}{10 \ }

f=20 \ s^{-1}

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f= 20 \ Hz

The frequency of the wave is <u>20 Hertz</u>

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Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
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Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

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Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

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Ilia_Sergeevich [38]
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4 years ago
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