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Gekata [30.6K]
3 years ago
12

A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal

force F⃗ to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)
Physics
1 answer:
AlekseyPX3 years ago
3 0

Answer:

x ( t )  = F / 5.5  t²

Explanation:

acceleration dv / dt = F / 5.5

dv = F / 5.5  dt

integraring on both sides

v =  F / 5.5  t + c

when t = 0 , v = 0

c = 0

v =  F / 5.5  t

ds / dt  = F / 5.5  t

integrating on both sides

s = F / 5.5  t² + c₁

when t = 0 , s = 0

c = 0

so

x ( t )  = F / 5.5  t²

This is the required relation.

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A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl
Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon m_1 and the mass of the basket m_2, then according to newton's second law:

(1). \:F = (m_1+m_2)a,

where a =0.265m/s^2 is the upward acceleration, and F = 688N is the net propelling force (counts the gravitational force).

Also, the tension T in the rope is 79.8 N more than the basket's weight; therefore,

(2). \:T = m_2g+79.8

and this tension must equal

T -m_2g =m_2a

(3). \:T = m_2g +m_2a

Combining equations (2) and (3) we get:

m_2a = 79.8

since a =0.265m/s^2, we have

\boxed{m_2 = 301.13kg}

Putting this into equation (1) and substituting the numerical values of F and a, we get:

688N = (m_1+301.13kg)(0.265m/s^2)

\boxed{m_1 = 2295 kg}

Thus, the mass of the balloon and the basket is  2295 kg and 301 kg respectively.

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3 years ago
The deepest well from which water can be pumped, by creating a vacuum on the top end of the pipe
Ivenika [448]
Can you further elaborate this isn't making much sense my mans
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3 years ago
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
Suppose there is a large amount of (weakly interacting) dark matter between us and a distant galaxy. How will this affect our vi
leonid [27]

Answer:

Dark matter does not affect our view, humans can see through them.

Explanation:

They do not affect our view because we can see right through the (weakly interacting) dark matter, as they do not interact or interfere with electromagnetic force.

Dark matter are often invisible substances and are difficult to spot because they don't absorb or reflect light.

7 0
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If you ride your bike 20 miles and it takes you 120 minutes,what is your average speed
Alexus [3.1K]
120 minutes=2 hours
20/2= 10mph
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3 years ago
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