determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.

Answers

Answer 1
Answer:

Answer: i think you should place it on the red line

Explanation:

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"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

Answers

Answer:

0.5 lambda(wavelength)

Explanation:

We know that

The first harmonic for both side open ended pipe is

L= 1/2lambda

So L = 0.5*wavelength

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?

Answers

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

Final answer:

Positive, negative, and zero terms in the energy equation. Calculation of energy output and metabolic power. Comparison of metabolic energy and power for different time durations.

Explanation:

To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:

  1. Positive terms:
  • ΔUg - the change in gravitational potential energy is positive as the barbell is lifted vertically from the ground.
  • ΔUs - the change in elastic potential energy is positive if there is any stretch or compression in the system.
Negative terms:
  • ΔK - the change in kinetic energy is negative as the barbell is lifted at a constant speed, so there is no change in velocity.
  • ΔEch - the change in chemical potential energy is negative if the athlete is not ingesting any food or drinks during the exercise.
Zero terms:
  • ΔEth - the change in thermal energy is zero if there is no heat transfer in the system.

To determine the energy output by the athlete, we can calculate the work done on the barbell using the formula W = ΔUg. In this case, the work done is equal to the change in gravitational potential energy, which is equal to mgh. Thus, W = 400 N × 2.0 m = 800 J. So the energy output by the athlete is 800 J.

The metabolic power can be calculated using the equation P = W / t, where P is the power, W is the work done, and t is the time taken. Substituting the given values, P = 800 J / 1.6 s = 500 W. Therefore, the metabolic power of the athlete is 500 W. If the task is performed in a faster time, the metabolic energy expended will be the same. However, the metabolic power will be greater as the work is done in less time.

Learn more about Energy equation here:

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Name the four forces in physics?​

Answers

Answer:

Gravitational

Electrostatic

magnetic

Frictional

gravitational

electrostatic

magnetic

frictional

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If half the kinetic energy of the initially moving object (m1m1) is transferred to the other object (m2m2), what is the ratio of their masses? Express your answers using three significant figures separated by a comma.

Answers

Answer:

0.25( m1m1) , 0.75( m2m2)

Explanation:

Noted the formula for kinetic energy is 1/2(Mass × Velocity).

Therefore the original value of the mass is 0.5, giving half away makes it 0.25 to another mass which is primarily 0.5. This now makes the new mass 0.25+0.5=0.75.

Thank you.

How does an increase in cold working effect Modulus of Elasticity and why?

Answers

Answer:

There is a decrease in modulus of elasticity

Explanation:

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.

Calculate the de Broglie wavelength of an electron and a one-ton car, both moving with speed of 100 km/hour. Based on your calculation could you predict which will behave like a "quantum particle" and why. Please explain each step in words and detail.

Answers

Answer :

(a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is 2.38*10^(-38)\ m

Explanation :

Given that,

Speed = 100 km/hr

Mass of car = 1 ton

(a). We need to calculate the wavelength of electron

Using formula of wavelength

\lambda_(e)=(h)/(p)

\lambda_(e)=(h)/(mv)

Put the value into the formula

\lambda_(e)=(6.63*10^(-34))/(9.1*10^(-31)*100*(5)/(18))

\lambda=0.00002622

\lambda=26.22*10^(-6)\ m

\lambda=26.22\ \mu m

(II).  We need to calculate the wavelength of car

Using formula of wavelength again

\lambda_(e)=(6.63*10^(-34))/(1000*100*(5)/(18))

\lambda=2.38*10^(-38)\ m

The wavelength of the electron is greater than the dimension of electron and the wavelength of car is less than the dimension of car.

Therefore, electron is quantum particle and car is classical.

Hence, (a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is 2.38*10^(-38)\ m.