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4 years ago

Plzzzz helppppppppppp

Physics

1 answer:

pashok25 [27]4 years ago

7 0

1. Explain, using Kinetic theory, why when you have a hot room, the room cools down when you open a window to the cold outside.

Answer:

According to kinetic theory of gases, at higher temperature, the molecules in air have higher kinetic energy and they are moving at a faster speed. Hence the pressure is greater. And Air moves from higher pressure to lower pressure.

In a hot room, the air pressure is greater. The cool air outside has lower pressure. On opening the window, Air moves from higher pressure to lower pressure, from outside to inside.

2. If an atom is 1*10^(-10) m wide, and the nucleus is 1*10^(-15) m wide, how wide would the atom be if the nucleus was scaled up to be 1 cm wide?

Answer:

When nucleus was scaled upto 1 cm, atom would be 10^(5) cm.

The question is pretty simple, we just have to scale the size of nucleus to larger value and get the size of atom accordingly.

We know that

when nucleus is 10^(-15) m => atom is 10^(-10) m

When nucleus becomes 1 m => atom is 10^(-10) * 10^(15) m = 10^(5) m

When nucleus is 1 * 10^(-2) => atom is 10^(5) * 10^(-2) m

When nucleus is 1 cm => atom is 10^(5) cm

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A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then t

lapo4ka [179]

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_{atm}$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}$

Where:

$P_{1}$ , $P_{2}$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the tank and ground level, measured in meters.

Given that $P_{1} = P_{2} = P_{atm}$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 0\,\frac{m}{s}$ , $z_{1} = 6.9\,m$ and $z_{2} = 4.9\,m$ , the expression is reduced to this:

$\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)$

And final speed is now calculated after clearing it:

$v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}$

$v_{2} \approx 6.263\,\frac{m}{s}$

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

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