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Trava [24]
3 years ago
11

A cannonball is launched off a 100 m cliff horizontally with an initial velocity of 50 m/s. The goal is to have the cannonball t

ravel a target distance of 400m. Based on this, will the cannonball reach the target distance?
a)

Yes, the cannonball will hit its goal


b)

No, the cannonball will overshoot its goal


c)

No, the cannonball will fall short of the goal


d)

More information is needed to determine the outcome
Physics
1 answer:
Rudiy273 years ago
7 0

Answer: It will fall short of its goal.

Explanation: I took the quiz.

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Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su
dybincka [34]
  1. La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
  2. La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua (v), en metros por segundo:

v = \sqrt{\frac{K}{\rho} } (1)

Donde:

  • K - Módulo de compresibilidad, en newtons por metro cuadrado.
  • \rho - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que \rho = 1\times 10^{3}\,\frac{kg}{m^{3}} y K = 2,16\times 10^{9}\,\frac{N}{m^{2}}, entonces la velocidad de las ondas sonoras es:

v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }

v\approx 1469,694\,\frac{m}{s}

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda (\lambda), en metros, mediante la siguiente fórmula:

\lambda = \frac{v}{f} (2)

Donde f es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que v\approx 1469,694\,\frac{m}{s} y f = 1000\,hz, entonces la longitud de onda de las ondas sonoras es:

\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}

\lambda = 1,470\,m

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

6 0
2 years ago
A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca
4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s  m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t +  1/2 a t²

s =  1/2 a t²   ----------- (2)

compare equation (1)  and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

3 0
3 years ago
Anna pushes a box with a force of 8.00 newtons. She generates a power of 3.00 watts. How much time does it take for Anna to move
QveST [7]
Power is the energy in a system per time.  It will have units of Watts which is equal to joules per second. It can be expressed as:

P = E / t

where E = Force x distance

P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s
8 0
4 years ago
Read 2 more answers
0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distanc
weqwewe [10]

Answer:

1.6 J

Explanation:

Work = change in energy

W = ΔKE

Fd = KE

(0.8 N) (2 m) = KE

KE = 1.6 J

4 0
3 years ago
how could you prove to a skeptic that the beats are an interference effect that requires both sound sources? test out your metho
podryga [215]

Beats are interference patterns between two tones of different frequencies. To prove the skeptic first, play the recorded audio as there are no beats in it. Now take two sound sources with different frequencies. When both sources are turned on, we hear notes that rise and fall at equal intervals. That's what's called beats.

A frequency beat occurs when two waves with different frequencies overlap, causing alternating cycles of constructive and destructive interference between the waves.

When we tap the table with our finger, then put our ear to the table, and tap the table surface as far as 30 cm from our ear. Then the sound of beats on the table will sound louder when we put our ears on the table. So, it can be concluded that solid objects can conduct sound better than air. This is because the molecules or particles of solid objects are denser than air.

Learn more about the beat's frequency at brainly.com/question/14157895

#SPJ4

7 0
1 year ago
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