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stira [4]
3 years ago
6

If a block of copper measures 2.00 cm x 4.00 cm x 5.00 cm and weighs 356 grams, what is its density?

Physics
1 answer:
soldier1979 [14.2K]3 years ago
3 0
Density = g/mL
cm^3 = mL

2*4*5 = 40 mL

density = 356/40 = 8.9 g/mL
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By the law of reflection which says that the angle of reflection is equal to the angle of incidence, which in other words says, sin x = sin y, it can be said, based on the given that the angle of reflection is also equal to the angle of incidence of 35 degrees
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3 years ago
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A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i
nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

U=mgh

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J

2. 41.1 J

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U=mgh

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J

3. 0 J

As before, the gravitational potential energy of the ball is given by

U=mgh

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

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4 years ago
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3 years ago
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Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air (\rho), measured in kilograms per cubic meter, is defined by following equation:

\rho = \frac{P\cdot M}{R_{u}\cdot T} (1)

Where:

P - Pressure, measured in kilopascals.

M - Molar mass, measured in kilomoles per kilogram.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

T - Temperature, measured in Kelvin.

If we know that P = 2026.5\,kPa, M = 28.965\,\frac{kg}{kmol}, R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} and T = 300\,K, then the density of air is:

\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}

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m = \rho \cdot V (2)

Where m is the mass, measured in kilograms.

If we know that \rho = 23.534\,\frac{kg}{m^{3}} and V = 10\,m^{3}, then the mass of air stored in the vessel is:

m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})

m = 235.34\,kg

The mass of air stored in the vessel is 235.34 kilograms.

7 0
3 years ago
gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an
Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

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Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

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PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0
3 years ago
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