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What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?

djyliett [7]

As you mentioned, we will use <span>Equipartition Theorem.
</span><span>H2 has 5 degrees of freedom; 3 translations and 2 rotation
</span>Therefore:
Internal energy = (5/2) nRT
You just substitute in the equation with the values of R and T and calculate the internal energy as follows:
Internal energy = (5/2) x 2 x <span>8.314 x 308 = 32.0089 x 10^3 J</span>

4 0

3 years ago

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

castortr0y [4]

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>

Let's draw the free body diagram of the system.
To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.

$F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\$

To find the answer, we have to find the tension,

$Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N$

Thus, the vertical component of the force exerted by the hi.nge on the beam will be,

$F_V=(29*9.8)-(169.43*sin57)=142.10N$

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

Learn more about the tension here:

brainly.com/question/28106868

#SPJ1

5 0

2 years ago

Read 2 more answers

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

Semenov [28]

Answer:

$f_n=3.75N$

Explanation:

From the question we are told that:

Frictional force $F=0.150N$

Coefficient of kinetic friction $\mu=0.04$

Generally the equation for Normal for is mathematically given by

$f_n=\frac{F}{\mu}$

Therefore

$f_n=\frac{0.150}{0.04}$

$f_n=3.75N$

5 0

3 years ago

A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a

ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω / t

(ω = √g/R)

∝ = $\frac{1}{t } \sqrt{\frac{g}{R} } }$

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = $\frac{2FR}{MR^2 + 2mr^2}$

Now we have ,

∝ = $\frac{1}{t} \sqrt{\frac{g}{R} }$ , ∝ = $\frac{2FR}{MR^2+2mr^2}$

equating both values

We have,

$F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }$

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

$F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N$

Each thruster has to applied a force of 294.5N in tangential direction

3 0

4 years ago

Analyze how some insects are able to move around on the surface of a lake or pond

koban [17]

It is called surface tension it is the elastic personality of some liquids as they pull together to take up as little surface area as possible. the water molecules would rather stay together than be pulled apart<span />

3 0

3 years ago