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4 years ago

Changing the length of an air column will alter its

Physics

2 answers:

netineya [11]4 years ago

8 0

The change in the length of air column alters the natural frequency.

Explanation:

Various resonance is obtained by changing the length of air column.

The length of air column is in inverse relation with natural frequency, that is, when air column length is increased, the natural frequency decreases. And, when air column length is decreased, the natural frequency is increased. The source frequency will not change without physically altering the source.

AveGali [126]4 years ago

7 0

Answer:

i think its pressure

Explanation:

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A ladybug sits 10.8 cm from the center of a beatles music album spinning at 33.33 rpm. What is the maximum velocity (in m/s) of

snow_lady [41]

Answer:

The maximum velocity is 0.377 m/s

Explanation:

Please, the solution is in the Word file attached

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3 years ago

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Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

oee [108]

Answer:

3.25 × 10^7 m/s

Explanation:

Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)

Ek = 1/2 mv2 = qΔV .................. 1

Given that V is the electron speed in m/s

Charge of electron = 1.60217662 × 10-19 coulombs

Mass of electron = 9.109×10−31 kilograms

ΔV = 3.0kV = 3000V

Make V the subject of the formula in eqaution 1

V = sqr root 2qΔV/m

V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31

V = 3.25 × 10^7 m/s

3 0

4 years ago

The image is the question

Evgen [1.6K]

Answer:

c is the answer because it i in a series not a parallel circut.

Explanation:

3 0

4 years ago

In a circus act, a 70 kg clown is shot from a cannon with an initial velocity of 17 m/s at some unknown angle above the horizont

EleoNora [17]

Answer:

$K_{2}=7302.4J$

Explanation:

Given the initial velocity of the clown, his mass and final height we can calculate the final kinetic energy using the conservation of total mechanical energy

$K_{1}+U_{1}=K_{2}+U_{2}$

$K_{2}=K_{1}+U_{1}-U_{2}$

$K_{2}=\frac{1}{2}mv_{1}^{2}+mgh_{1}-mgh_{2}$

Since $h_{1}=0$

$K_{2}=\frac{1}{2}mv_{1}^{2}-mgh_{2}$

$K_{2}=\frac{1}{2}(70kg)(17m/s)^2-(70kg)(9.8m/s^2)(4.1m)=7302.4J$

$K_{2}=7302.4J$

7 0

3 years ago

An object in space has altitude of 2210 km, velocity of 7000 m/s and flight path angle of 20 degrees. Find the eccentricity, sem

Dmitry [639]

Answer:

Eccentricity = 0.0557

Semi-major axis = 9,095 km

Angular momentum/mass = 60,116 $\frac{km^{2} }{Sec}$

Kinetic energies/mass = 24,500 KJ

Potential energies/mass = 21,673 KJ

Explanation:

Eccentricity

To find the eccentricity use the following formula

Eccentricity = [ Altitude from the earth x ( $Velocity^{2}$ / Gravitational parameter for the Earth ) ] - 1

Where

Altitude from the earth radius = Radius of the earth + Altitude of the earth from radius = 6,378 km + 2,210 km = 8,588 km

Velocity = 7,000 m/s

Gravitational parameter for the Earth = 3.986004418 × $10^{14}$

Eccentricity = ?

Placing values in the formula

Eccentricity = [ ( 8,588 km x 1000 ) x ( $7000^{2}$ / 3.986004418 × $10^{14}$ ) ] - 1

Eccentricity = 0.0557

Semi-major axis

Total Distance = Semi-major axis x ( 1 - Eccentricity )

Where

Total Distance = 8,588 km

Eccentricity = 0.0557

8,588 x 1,000 m = Semi-major axis x ( 1 - 0.0557 )

8,588,000 m = Semi-major axis x 0.9443

Semi-major axis = 8,588,000 m / 0.9443

Semi-major axis = 9,094,567.40 m

Semi-major axis = 9,094.56740 km

Semi-major axis = 9,095 km

Angular momentum/mass

L = MVR

L/M = VR

Where

V = Velocity = $\frac{7,000 m/s}{1000}$ = 7 km/s

R = Total Radius = Radius of Earth + Altitude = 6,378 km + 2,210 km = 8,588 km

Placing values in the formula

L/M = 7 km/s x 8,588 km = 60,116 $\frac{km^{2} }{Sec}$

Kinetic energies/mass

Ke = I $W^{2}$

Ke = $\frac{1}{2} mr^{2}$ $W^{2}$ ( Where I = $mr^{2}$ )

$\frac{ke}{m}$ = $\frac{r^2}{2}$ $W^{2}$

$\frac{ke}{m}$ = $\frac{r^2}{2}$ $\frac{V^2}{r^2}$ ( $W^{2}$ = $\frac{V^2}{r^2}$ )

$\frac{ke}{m}$ = $\frac{V^2}{2}$

$\frac{ke}{m}$ = $\frac{7000^2}{2}$

$\frac{ke}{m}$ = 24,500,000 J

$\frac{ke}{m}$ = 24,500 KJ

Potential energies/mass

PE = mgh

$\frac{PE}{m}$ = gh

Where

g = 9.807 $\frac{m}{s^2}$

h = 2,210 km x 1,000 = 2,210,000 m

Placing values in the formula

$\frac{PE}{m}$ = 9.807 $\frac{m}{s^2}$ x 2,210,000 m

$\frac{PE}{m}$ = 21,673,470 J

$\frac{PE}{m}$ = 21,673.470 KJ

$\frac{PE}{m}$ = 21,673 KJ

8 0

3 years ago