The answer is true: the pressure of a gas will decrease as temperature decreases in a rigid container.

This is one of the central gas laws called the Gay-Lussac law that states for a given gas at a constant volume, the pressure of the gas is directly proportional to its temperature. We also know that as temperature reduces, so too does molecular interaction. Increased temperature results in increased pressure, and decreased temperature therefore results in decreased pressure.

6 0

3 years ago

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You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type

alexdok [17]

Answer:

$30 N \cdot m$

Explanation:

The torque applied by a force can be calculated as

$\tau = F d sin \theta$

where

F is the magnitude of the force

d is the length of the arm

$\theta$ is the angle between the direction of the force and the arm

In this problem, we have

F = 15 N

d = 2.0 m

$\theta=90^{\circ}$

Substituting into the equation, we find

$\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m$

7 0

3 years ago

A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring

sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0

3 years ago