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3 years ago

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Titan Tommy and the Test Tubes at a nightclub this weekend. The lead instrumentalist uses a test tube (closed-end air column) wi

th a 17.2 cm air column. The speed of sound in the test tube is 340 m/sec. Find the frequency of the first harmonic played by this instrument.

1 answer:

Westkost [7]3 years ago

6 0

Hello , it' s 5 but i am not sure Explanation:

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True or false<br><br>energy can neither be created nor be destroyed<br>

muminat

Answer:

the correct answer is TRUE

Hope it helps you

have a nice day

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3 years ago

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how do you Compute the approximate acceleration of gravity for an object above the earth's surface, assigning accel gravity with

solong [7]

The approximate acceleration of gravity for an object above the earth's surface is 9.8 m/s².

<h3>Acceleration due to gravity </h3>

The acceleration due to gravity of an object above the Earth's surface is calculated from Newton's second law of motion and Newton's law of universal gravitation.

<h3>Force between the object and Earth</h3>

$F_g = \frac{GmM}{R^2}$

<h3>Weight of the object</h3>

$F_g = mg\\$

Solve (1) and (2)

mg = GmM/R²

g = GM/R²

where;

M is mass of Earth
G is gravitational constant
R is radius of the Earth = 6,371 km

g = (6.673 x 10⁻¹¹ x 5.98 x 10²⁴)/(6,371 x 10³)²

g = 9.8 m/s²

Learn more about acceleration due to gravity here: brainly.com/question/88039

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2 years ago

Dos Autos se acercan uno hacia el otro como se muestra en la figura. Si los carros chocan y quedan pegados, DETERMINE: a) las co

lisov135 [29]

Answer:

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Explanation:

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3 years ago

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and

Zigmanuir [339]

Answer:

$F=5.8\times 10^{8}\ N$

$F=35.57\times 10^{21}\ N$

Explanation:

Given that

Intensity I

$I= 1.39\ KW/m^2$

$Speed\ of \ light = 2.99\times 10^8\ m/s$

Radius of earth,R = 6370 Km

We know that surface area of earth, A

$A=\pi R^2$

$A=\pi (6370\times 10^3)^2\ m^2$

$A=1.27\times 10^{14}\ m^2$

As we know that pressure due to intensity given as

$P=\dfrac{I}{V}$

V =Velocity of light

$V=3\times 10^8\ m/s$

$P=\dfrac{1.39\times 1000}{=3\times 10^8}$

$P=4.6\times 10^{-6}\ Pa$

We know that force F

F = P .A

$F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N$

$F=5.8\times 10^{8}\ N$

b)Gravitational force F

$F=G\dfrac{m.M}{r^2}$

$M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg$

$r =1.5\times 10^{11}\ m$

$G =6.67\times 10^{-11}Nm^2/kg^2$

So F

$F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}$

$F=35.57\times 10^{21}\ N$

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3 years ago

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Jake drove 160 kilometres in 2 hours. What was his speed, in kmh-1

ruslelena [56]

Answer:

80 kmh

Explanation:

IDK lol i just divided it by 2 because he drove 80 kilometres in one hour

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3 years ago

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