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2 years ago

11

At what temperature would volume of gas be doubled of pressure at the same time increase from 700 to 800 mm of Hg? please answer

me fast

1 answer:

Gnoma [55]2 years ago

8 0

Answer:

P1V1 = nRT1

P2V2 = nRT2

Divide one by the other:

P1V1/P2V2 = nRT1/nRT2

From which:

P1V1/P2V2 = T1/T2

(Or P1V1 = P2V2 under isothermal conditions)

Inverting and isolating T2 (final temp)

(P2V2/P1V1)T1 = T2 (Temp in K).

Now P1/P2 = 1

V1/V2 = 1/2

T1 = 273 K, the initial temp.

Therefore, inserting these values into above:

2 x 273 K = T2 = 546 K, or 273 C.

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if the current in a wire is 2.0 amperes and the potential difference across the wire is 10 volts what is the resistance of the w

Pavlova-9 [17]

Answer:

R = 2Ω

Explanation:

Potential difference (V) = current (I) * Resistance (R)

V = IR

I = 2.0A

V = 10v

R = ?

V = IR

R = V / I

R = 10 / 2

R = 2Ω

The resistance across the wire is 2Ω

3 0

3 years ago

Read 2 more answers

Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha

Reptile [31]

Answer:

(i) W = 8.918 N

(ii) $V = 9.1 \times 10^{-4} m^3$

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

$W = mg$

$W = \rho V g$

here we know that

$\rho = 0.91 g/cm^3$

$Volume = L^3$

$Volume = 10^3 = 1000 cm^3$

now the mass of the ice cube is given as

$m = 0.91 \times 1000 = 910 g$

now weight is given as

$W = 0.910 \times 9.8 = 8.918 N$

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

$W_{displaced} = 8.918 N$

So here volume displaced is given as

$\rho_{water}Vg = 8.918$

$1000(V)9.8 = 8.918$

$V = 9.1 \times 10^{-4} m^3$

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

$W = \rho_{water} (L^2 d) g$

$8.918 = 1000(0.10^2 \times d) 9.8$

$d = 0.091 m$

so it is submerged by d = 9.1 cm inside water

4 0

3 years ago

A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface. The surface of the ramp is the x-

kifflom [539]

Magnitude of the force of tension: 139 N

Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

The force of tension, $F_T$ , acting up along the plane
The force of friction, $F_f=14.8 N$ , acting down along the plane
The component of the weight in the x-direction, $F_{gx}$ , acting down along the plane

We know that the magnitude of the weight is

$F_g=70.0 N$

So its x-component is

$F_{gx}=F_g sin \theta =(70.0)(sin 22^{\circ})=26.2 N$

The net force along the x-direction can be written as

$F_x = F_T-F_f-F_{gx}$

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

$F_T=F_x+F_f+F_{gx}=98+14.8+26.2=139 N$

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0

3 years ago

Read 2 more answers

A mine elevator is supported by a single steel cable 0.0125 m in diameter. The total mass of the elevator cage and occupants is

Sonja [21]

Answer:

0.8895m

Explanation:

Cable diameter = 0.0125m

Mass of elevator = 6450kg

Young Modulus(E) = 2.11*10¹¹N/m

∇l (change in length) =

L = 362m

A = Πr², but r = d / 2 = 0.0125 / 2 = 0.00625m

A = 3.142 * (0.00625)² = 1.227*10^-4m²

Young Modulus (E) = Tensile stress / Tensile strain

E = (F / A) / ∇l / L

F = mg = 6450 * 9.8 = 63210N

2.11*10¹¹ = (63210 / 1.22*10^-4) / (∇l / 362)

2.11*10¹¹ = 5.18*10⁸ / (∇l / 362)

2.11*10¹¹ = (5.18*10⁸ * 362) / ∇l

2.11*10¹¹ = 1.875*10¹¹ / ∇l

∇l = 1.875*10¹¹ / 2.11*10¹¹

∇l = 0.8895m

The change in length is 0.8895m

8 0

3 years ago

When we go up From the earth surface.what happens about the atmospheric pressure

topjm [15]

Answer:

There are two reasons why air pressure decreases as altitude increases: density and depth of the atmosphere. Most gas molecules in the atmosphere are pulled close to Earth's surface by gravity, so gas particles are denser near the surface.

Explanation:

7 0

3 years ago