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3 years ago

How many MPH does an object have to go in order to break the sound barrier

1 answer:

5 0

Well, it's kind of a slippery number, because the speed of sound
depends on the air that the sound is traveling through, so it's actually
different for different aircraft flying at different altitudes, where the
pressure, temperature, and density of the air are different.

Here are some "standard" values"

On the ground . . . 761 mph

10,000 feet 733 mph

20,000 feet 705 mph

30,000 feet 677 mph

50,000 feet 660 mph

100,000 feet 676 mph

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Un montañero de 65kg de masa ha ascendido a la cima del Everest, la montaña más alta del mundo de 8848m de altura sobre el nivel

andrew-mc [135]

Answer:

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

Explanation:

Por la definición de trabajo sabemos que el montañero debió contrarrestar trabajo causado por la gravedad terrestre. Si asumimos que el cambio de la altura es muy pequeño en comparación con el radio del planeta (6371 kilómetros vs. 0,5 kilómetros), entonces podemos considerar que la aceleración gravitacional es constante y la ecuación de trabajo ( $\Delta W$ ), medido en joules, que reducida a:

$\Delta W = m\cdot g\cdot \Delta z$ (1)

Donde:

$m$ - Masa del montañero, medido en kilogramos.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$\Delta z$ - Distancia vertical de ascenso del montañero, medida en metros.

Si tenemos que $m = 65\,kg$ , $g = 9,807\,\frac{m}{s^{2}}$ y $\Delta z = 500\,m$ , entonces el trabajo realizado por el montañero para subir es:

$\Delta W = (65\,kg)\cdot \left(9,807\,\frac{m}{s^{2}} \right)\cdot (500\,m)$

$\Delta W = 318727,5\,J$

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

7 0

3 years ago

The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee

kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2> $v_f = 35.3 m/s$ </h2>

Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$

$C_d = 0.35$

$A = \frac{\pi d^2}{4}$

$A = \frac{\pi(0.074)^2}{4}$

$A = 4.3 \times 10^{-3} m^2$

$v = 40.2 m/s$

so we have

$F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}$

$F = 1.46 N$

So acceleration of the ball is

$a = \frac{F}{m}$

$a = \frac{1.46}{0.145}$

$a = 10.1 m/s^2$

Part B)

As per kinematics we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 40.2^2 = 2(-10.1)(18.4)$

$v_f = 35.3 m/s$

4 0

4 years ago

A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875

mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

5 0

3 years ago

Read 2 more answers

Student Exploration: Nuclear Decay. Has anyone done a Gizmos lab on this?

LUCKY_DIMON [66]

Answer:NOOPE

Explanation:IM THE MYSTERY MAN WHOOOOSHHHHHHHHHHHHHHH ???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

6 0

3 years ago

A boy weighs 40 kilograms. He runs at a velocity of 4 meters per second north. Which is his momentum?

Bogdan [553]

The boy's momentum is 160 kg*m/s north.

The formula of momentum is p = mv, where p is momentum.
p = 40 kg * 4m/s north
p =160 kg*m/s north

<span>Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.</span>

4 0

3 years ago

Read 2 more answers