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Oksi-84 [34.3K]
3 years ago
13

If pressure of a gas is increased in its volume at constant what will happen to its temperature

Physics
2 answers:
mrs_skeptik [129]3 years ago
8 0
The temperature will increase as well
tresset_1 [31]3 years ago
5 0

Answer:

Increase

Explanation:

If we assume the gas is ideal, then we can use the ideal gas law:

PV = nRT

If V is held constant while P increases, then T will also increase.

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I shared a picture of the problem. It’s a basic Physics question and an Algebra question.
julia-pushkina [17]

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

Given the expressions;

T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}

Equating both expressions we will have;

2 \pi \sqrt{\frac{m}{k} }  = \frac{2 \pi}{\omega}

Divide both equations by 2π

\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\

Square both sides

(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}

Take the square root of both sides

\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

3 0
3 years ago
A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.
pav-90 [236]
<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

           Displacement, s = 0 m

           Acceleration, a = -9.81 m/s²

            Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

                s = ut + 0.5 at²

                0 = 128.89 x t + 0.5 x (-9.81) x t²

                t² - 26.28 t = 0

                t ( t- 26.28) = 0

               t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

5 0
3 years ago
I need help ASAP please :)
rusak2 [61]

Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).

M = V d

M = 1.4 * 2 = 2.8 kg

7 0
2 years ago
When sound travels through air, the particles _____. a. vibrate along the direction the wave travels b. vibrate but not in any f
Mandarinka [93]
Vibrate along the direction the wave travels
8 0
3 years ago
If a snowboarder’s initial speed is 4 m/s and comes to rest when making it to the upper level. With a slightly greater initial s
Brrunno [24]

(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).

(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.

<h3>Conservation of mechanical energy</h3>

The effect of height  and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;

ΔK.E = ΔP.E

¹/₂m(v²- u²) = mg(hi - hf)

¹/₂(v²- u²) = g(0 - hf)

v² - u² = -2ghf

v² = u² - 2ghf

where;

  • v is the final velocity at upper level
  • u is the initial velocity
  • hf is final height
  • g is acceleration due to gravity

when u² = 2gh, then v² = 0,

when gravity reduces, u² > 2gh, and v² > 0

Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).

<h3>Final speed</h3>

v² = u² - 2ghf

where;

  • u is the initial speed = 5 m/s
  • g is acceleration due to gravity and its less than 9.8 m/s²
  • v is final speed
  • hf is equal height

Since g on Epislon is less than 9.8 m/s² of Earth;

5² - 2ghf > 3 m/s

Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.

Learn more about conservation of mechanical energy here: brainly.com/question/6852965

5 0
2 years ago
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