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GarryVolchara [31]

4 years ago

8

A gas with volume of 3 L and a pressure of 1 atm is moved to a new environment where the gas’s volume decreases to 1.50 L. What

is the new pressure on atm assuming temperature is held constant

1 answer:

Phoenix [80]4 years ago

3 0

Answer:

2 atm

Explanation:

Ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is gas constant,

and T is absolute temperature.

If n, R, and T are held constant:

PV = PV

(1 atm) (3 L) = P (1.50 L)

P = 2 atm

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Define refraction of light

Elza [17]

Answer:

Refraction is the slowing down of light waves as they pass into a different medium and speeding as they pass out.

hope it helps. plz mark me as brainliest.

6 0

3 years ago

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a consta

bonufazy [111]

Answer:

acceleration = 22.14 m/s²

time = 2.80 s

Explanation:

given data

initial velocity = 31.0

final velocity =

time = 1.40

solution

we get here acceleration that is

acceleration = $\frac{v(final) - v(initial)}{t}$ ...............1

acceleration = $\frac{31-0}{1.40}$

acceleration = 22.14 m/s²

and

by kinematic equation of motion

v = u +at .........2

t = $\frac{62-0}{22.14}$

time = 2.80 s

8 0

3 years ago

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.

5 0

4 years ago

Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for l

liberstina [14]

Explanation:

Given: $\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}$

$\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}$

$\:\:\:\:\:= 6×10^{14}\:\text{Hz}$

The work function $\phi$ is then

$\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})$

$\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}$

3 0

3 years ago

Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M an

Mumz [18]

Answer:

The hoop

Explanation:

We need to define the moment of inertia of the different objects, that is,

DISK:

$I_{disk} = \frac{1}{2} mR^2$

HOOP:

$I_{hoop} = mR^2$

SOLID SPHERE:

$I_{ss} = \frac{2}{5}mR^2$

HOLLOW SPHERE

$I_{hs} = \frac{2}{3}mR^2$

If we have the same acceleration for a Torque applied, then

$mR^2>\frac{2}{3}mR^2>\frac{1}{2} mR^2>\frac{2}{5}mR^2$

$I_{hoop}>I_{hs} >I_{disk}>I_{ss}$

The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration

4 0

3 years ago