The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?

Answers

Answer 1

Answer:

Answer:

a. 409.5 m/s b. f₁ = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz

Explanation:

a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5

f₃ = 5v/4L

= 5 × 409.5 m/s ÷ (4 × 0.75 m)

= 5 × 409,5 m/s ÷ 3 m

= 682.5 Hz

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A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.50 m/s and observes that it takes 1.2 s to reach the water. How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Round your answer to the nearest whole number.

Answers

Answer:

The answer is 9 m.

Explanation:

Using the kinematic equation for an object in free fall:

$y = y_o - v_o-(1)/(2)gt^(2)$

In this case:

$v_o = \textrm{Initial velocity} = 1.5[m/s]\nt = \textrm{air time} = 1.2 [s]$ $y_o = 0$

$g = \textrm{gravity} = 9.8 [m/s^(2) ]$

Plugging those values into the previous equation:

$y = 0 - 1.5*1.2-(1)/(2)*9.8*1.2^(2) \ny = -8.85 [m] \approx -9 [m]$

The negative sign is because the reference taken. If I see everything from the rescuer point of view.

A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/s – (0.0180 m/s3)t^2 ]i+[2.00 m/s + (0.550 m/s2)t ]j .a) What are ax(t) and ay(t), the x- and y- components of cars acceleration as a function of time?
b) What are the magnitude and direction of the velocity of the car at t= 8 sec?
c) What is the magnitude and direction of cars acceleration at t=8 sec

Answers

Maybe try A for your answer

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979.(a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement?

(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?

(c) What was his total displacement relative to the air mass?

Answers

Answer:

a) $s=35794.2\ m$

b) $v_w=3.53\ m.s^(-1)$

c) $s_w=56074.2\ m$

Explanation:

Given:

duration of flight, $t=169* 60=10140\ s$

velocity of flight, $v=3.53\ m.s^(-1)$

direction of flight, $45^(\circ)$ to the south of east

Now the total displacement:

$s=v.t$

$s=3.53* 10140$

$s=35794.2\ m$

Velocity of air, $v_a=2\ m.s^(-1)$

When the aircraft encounters a headwind in the opposite direction to the velocity of motion then the speed of the aircraft is lowered with respect to the ground.

But when the speed is observed with respect to the wind the reduced velocity of the aircraft is observed from an opposite moving wind having a magnitude equal to the difference in velocity of the aircraft. This results in no change in the apparent velocity of the aircraft.

Mathematically:

Velocity of the aircraft with respect to the ground:

$v_(g)=v-v_a$

$v_(g)=3.53-2$

$v_g=1.53\ m.s^(-1)$

Now the velocity of the aircraft with respect to the wind:

$v_w=v_g+v_a$

$v_w=1.53+2$

$v_w=3.53\ m.s^(-1)$

Now the total displacement with respect to the wind:

$s_w=v_w.t+v_a.t$

$s_w=3.53* 10140+2* 10140$

$s_w=56074.2\ m$

A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter stick hangs a 5.202 kg concrete block. From the 75.0 cm mark on the meter stick hangs a 7.99 kg steel ball. At what mark on the meter stick must a 2.46 kg wooden block be attached so that the meter stick balances horizontally when lowered into fresh water? Assume the densities of concrete, steel, and wood are 2500.0 kg/m3, 8000.0 kg/m3, and 500 kg/m3 respectively. A. 57.6 cm
B. 57.4 cm
C. 57.2 cm
D. 57.8 cm

Answers

The wooden block must be attached at the 57.6 cm mark on the meter stick.

What is the value

We can solve this equation for the distance of the wooden block from the pivot point, which is the mark on the meter stick where it must be attached so that the meter stick balances horizontally when lowered into fresh water.

The density of aluminum is 2700 kg/m³ The volume of the meter stick is 0.01 m³ (length times width times thickness). The acceleration due to gravity is 9.81 m/s²

Substituting these values into the equation above, we get:

distance of wooden block from pivot point = -44.89 Nm / [(2700 kg/m³)(0.01 m³)(9.81 m/s²)]

= 0.576 m

Learn more about meter

brainly.com/question/1578784

#SPJ3

Answer:

Stop cheating in exam

Explanation:

Shame!!!!

I am sorry but I will have to refer you to the student conduct at UTA.

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

The volume flow rate is 3.27m³/s

Diameter at the refinery is 88.64cm

Explanation:

Given

At the wellhead

Pipes diameter, d2 = 59.1cm = 0.591m

Flow speed of petroleum f2 = 11.9m/s

At the refinery,

Pipes diameter, d1 = ? Unknown

Flow speed of petroleum, f1 = 5.29m/s

Calculating the volume flow rate of petroleum along the pipe.

Volume flow rate = Flow rate * Area along the pipe

V = 11.9 * πd²/4

V = 11.9 * 22/7 * 0.591²/4

V = 3.265778m³/s

The volume flow rate is 3.27m³/s -------- Approximated

Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...

Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends

This gives;

V1A1 = V1A2

V1*πd1²/4 = V2 * πd2²/4 ----------- Divide through by π/4

So, we are left with

V1d1² = V2d2²

5.29 * d1²= 11.9 * 59.1²

d1² = 11.9 * 59.1²/5.29

d1² = 7857.172

d1 = √7857.172

d1 = 88.6406904305240618

d1 = 88.64cm --------------- Approximated

Answer all three parts and show work.

Answers

The distance for both Parts A and B are given in the question.

A balloon drifts 140m toward the west in 45s.

The wind suddenly changes and the balloon flies 90m toward the east in the next 25s.

To find the total distance, we can just add.

140 + 90 = 230m

Best of Luck!

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