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pychu [463]
2 years ago
15

The pressure of a gas always increases with increasing temperature. A. True B. False​

Physics
2 answers:
olga2289 [7]2 years ago
7 0

True

Gay Lussac's Law

When the volume is constant, the gas pressure is proportional to its absolute temperature

\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

maks197457 [2]2 years ago
4 0

Answer:

True

Explanation:

This can be easily understood by visualising the particles of gas in the container moving with a greater energy when the temperature is increased. This means that they have more collisions with each other and the sides of the container and hence the pressure is increased.

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Compared to a stationary galaxy, light from a galaxy that is moving away from earth will appear _____.
Vlad1618 [11]
"B" When an object moves away from us, the light is shifted to the red end of the spectrum, as its wavelengths get longer.
3 0
3 years ago
Read 2 more answers
A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the
Mumz [18]

Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

F=\frac{2\times0.588}{\sqrt{3} }

F = 0.679 N

6 0
4 years ago
A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature
shepuryov [24]

You can make sure there's no change in volume by keeping
your gas in a sealed jar with no leaks.  Then you can play with
the temperature and the pressure all you want, and you'll know
that the volume is constant.

For 'ideal' gases,

       (pressure) times (volume) is proportional to (temperature).

And if volume is constant, then

                 (pressure) is proportional to (temperature) .

So if you increase the temperature from 110K to 235K,
the pressure increases to  (235/110)  of where it started.

     (400 kPa) x (235/110)  =  854.55 kPa. (rounded)

Obviously, choice-b is the right one, but
I don't know where the .46 came from.
 
4 0
3 years ago
Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is t
zlopas [31]

Answer:

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

Explanation:

As we know that the wave equation is given as

y = A sin(\omega t - k x + \phi_0)

now we have

A = 0.19 m

\lambda = 2.6 m

so we have

k = \frac{2\pi}{\lambda}

k = \frac{2\pi}{2.6}

k = 2.42  per m

also we have

T = 1.2 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{1.2}

\omega = 5.23 rad/s

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

\phi_0 = \frac{\pi}{2}

so we have

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

6 0
4 years ago
A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at
baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

\Sigma \tau = 0

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
\tau = r \times F

Doing the summation using their respective lever arms:

0 = L Tsin\theta  - dF_g

dF_g = LTsin\theta

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o

Now, let's solve for 'T'.

T = \frac{dMg}{Lsin\theta}

Plugging in the values:
T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}

3 0
2 years ago
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