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3 years ago

The speed of sound I gases depends on a humidity b air c temperature d visibility

Physics

1 answer:

Mazyrski [523]3 years ago

4 0

The speed of sound in gases is represented mathematically as:

v(sound) = √γRT/M

where γ = adiabatic constant which is the ratio of the heat capacities at constant pressure and constant volume; Cp/Cv

R = gas constant

T = temperature

M = molar mass of the gas

Hence, the speed of sound in gas is directly proportional to the square root of temperature.

Ans: c) temperature

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A ping-pong ball is shot straight north from a popgun at 4.0 m/s. The wind is blowing to the west at 3.0 m/s. What is the actual

Vlad1618 [11]

Actual velocity of the ping-pong ball= 5 m/s

Explanation:

velocity of ping pong ball because of the shot gun= 4 m/s North

velocity added to the ping-pong ball due to the wind=3 m/s

These velocities are perpendicular to each other. so we use Pythagoras theorem to find the resultant velocity of the ping- pong ball

so the actual velocity of the ping-pong ball =V= √4²+3²

V= √25

V= 5 m/s

5 0

4 years ago

A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force

Vlada [557]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec
vf = 0 m/sec 
Fμ = 7100 N (Force due to friction) 
Fg = 14700 N 
With the force due to gravity, you can find the mass of the car: 
F = ma 
14700 N = m (9.8 m/sec²) 
m = 1500 kg 
Now, we can use the equation again to find the deacceleration due to friction: 
F = ma 
7100 N = (1500 kg) a 
a = 4.73333333333 m/sec² 
And now, we can use a velocity formula to find the distance traveled: 
vf² = vo² + 2a∆d 
0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d 
0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d 
-625 m²/sec² = (-9.466666666667 m/sec²) ∆d 
∆d = 66.0211267605634 m 
∆d = 66.02 m

7 0

4 years ago

Short-term memory is active, while long-term memory is:

Afina-wow [57]

Answer:

the answer is b. reflective

6 0

3 years ago

An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance

Lorico [155]

Answer:

The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.

Explanation:

Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

$F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}$

and since the electric field E in between parallel plates separated a distance d and under a potential difference $\Delta V$ , is given by:

$E=\frac{\Delta\,V}{d}$

then :

$a=\frac{q\,\Delta V}{m\,d}$

We want to find when the particle reaches velocity zero via kinematics:

$v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a$

We replace this time (t) in the kinematic equation for the particle displacement:

$\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}$

Replacing the values with the information given, converting the distance d into meters (0.01 m), using $\Delta V=100\,V$ , and the electron's kinetic energy:

$\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J$

we get:

$\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters$ Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:

0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]

8 0

4 years ago

The density of dry air at room temperature and atmospheric pressure Is 1.2 kg m-3. What volume would 6 kg of dry air occupy unde

larisa [96]

Answer:

V = 5 m³

Explanation:

The density of air, d = 1.2 kg/m³

Mass of the dry air, m = 6 kg

We need to find the volume o the gas. We know that, the density of an object is given by mass divided by its volume. So,

$d=\dfrac{m}{V}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{6}{1.2}\\\\V=5\ m^3$

So, the volume of the dry air is 5 m³.

4 0

3 years ago