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bagirrra123 [75]
3 years ago
15

A severe thunderstorm dumped 2.4 in of rain in 30 min on a town of area 29 km2. What mass of water fell on the town? One cubic m

eter of water has a mass of 103 kg.
Physics
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:

The mass of water is 1.7661\times10^{9}\ kg

Explanation:

Given that,

Area of town = 29 km²

Time = 30 min

Height = 2.4 inch

Mass = 103 kg

We know that,

1 m = 39.37 inch

So, 2.4\ inch =\dfrac{2.4}{39.37}

2.4\ inch=0.0609\ m

We need to calculate the volume

Using formula of volume

V=H\times A

Put the value into the formula

V=0.0609\times29\times1000^2

V=1766100\ m^3

We need to calculate the mass of water

Using formula of density

\rho=\dfrac{m}{V}

Put the value into the formula

1000=\dfrac{m}{1766100}

m=1000\times1766100

m=1.7661\times10^{9}\ kg

Hence, The mass of water is 1.7661\times10^{9}\ kg

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stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

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Explanation:

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La respuesta es sí, hay una fuerza que actúa sobre el móvil A y es la única fuerza ya que A cae libremente bajo la influencia de la fuerza.

Explanation:

Según la primera ley de movimiento de Newton, un cuerpo continuará en su estado de reposo o en un movimiento uniforme en línea recta a menos que actúen sobre él fuerzas impresas.

Dado que el móvil A cae libremente, desde su estado de reposo inicial, según la primera ley de movimiento de Newton, experimenta una fuerza que actúa sobre él para hacer que caiga y continúe en caída libre.

El móvil B se mueve con una velocidad constante, por lo tanto, de acuerdo con la primera ley de movimiento de Newton, no hay fuerzas impresas que actúen sobre él.

El móvil C está completamente en reposo en el suelo, por lo tanto, tampoco hay fuerzas que actúen sobre él.

La respuesta es sí, hay una fuerza actuando sobre el móvil A y es la única fuerza cuando A cae libremente bajo la influencia de la fuerza.

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