The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At a specific temperature the pressure is 102.1 kPa at sea level and 87.8 kPa at h = 1,000 m. (Round your answers to one decimal place.) (a) What is the pressure at an altitude of 4500 m? kPa (b) What is the pressure at the top of a mountain that is 6165 m high?

Answers

Answer 1
Answer:

Final answer:

The rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. Using this information, we can calculate the pressure at different altitudes.

Explanation:

To solve this problem, we can use the fact that the rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. We can set up a proportion using the given information to find the constant of proportionality. Then, we can use this constant to find the pressure at different altitudes.

(a) Let's use the given information to find the constant of proportionality. We have P = kP, where k is the constant of proportionality. Using the values at sea level and 1000m, we can set up the proportion 102.1/87.8 = k. Solving for k, we find k ≈ 1.16.

Now, we can use this constant to find the pressure at an altitude of 4500m. We set up the proportion 102.1/x = 1.16, where x is the pressure at 4500m. Solving for x, we find x ≈ 122.0 kPa.

(b) We can use the same constant of proportionality to find the pressure at the top of a mountain that is 6165m high. We set up the proportion 102.1/x = 1.16, where x is the pressure at the top of the mountain. Solving for x, we find x ≈ 89.2 kPa.

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A trumpet player hears 5 beats per second when she plays a note and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. Was her initial frequency 435 Hz or 445 Hz? Explain.

Answers

Answer:

her initial frequency is 445 Hz

Explanation:

Given;

initial beat frequency, F_B = 5

observed frequency, F = 440 Hz

let the initial frequency = F₁

F₁ = F  ±  5 Hz

F₁ = 440 Hz  ±  5 Hz

F₁ = 435 or 445 Hz

This result obtained shows that her initial frequency can either be 435 Hz or 445 Hz

The last beat frequency will be used to determine the actual initial frequency.

F = v/λ

Frequency (F) is inversely proportional to wavelength. That is an increase in length will cause a proportional decrease in frequency.

This shows that the final frequency is smaller than the initial frequency because of the increase in length.

Initial frequency   -  frequency of tuning fork = 5 beat frequency

Reduced initial frequency - frequency of tuning fork = 3 beat frequency

Initial frequency = 5Hz + 440 Hz = 445 Hz

Final frequency (Reduced initial frequency) = 440 + 3 = 443 Hz

Check: 445 Hz - 440 Hz = 5 Hz

            443 Hz - 440 Hz = 3 Hz

A missile is moving 1350 m/s at a 25.0 angle

Answers

i will answer both versions assuming what you want to know is the distance it travels up from and over the ground. and how long until it reaches space. 540 meters per second up and over. to reach space which is 100km above sea level, it would take about 5400 minutes

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done by the force

Answers

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

\Rightarrow a=(F)/(m)

Hence acceleration of the block is given by:

\Rightarrow a=(25)/(2)=12.5m/s^2

Displacement of the object is given by:

\Rightarrow S=ut+(1)/(2)at^2

Substitute the values

\Rightarrow S=0*3+(1)/(2)(12.5)3^2

\Rightarrow S=56.25 m

Now work done is given by:

 W=F.S

W = 25×56.25

W =1406.25 J

A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?

Answers

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

Nitrogen makes up about what percent of a human's body weight?

Answers

Answer:

the answer is 3.3 %

Explanation:

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

Answers

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

Final answer:

To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Explanation:

The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

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