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crimeas [40]
2 years ago
14

What is the speed of sound in dry air at 0°C? A. 316 meters/second B. 331 meters/second C. 346 meters/second D. 373 meters/secon

d
Physics
2 answers:
Katena32 [7]2 years ago
8 0

Answer:

1,450 meters per second

Explanation:

amm18122 years ago
3 0

Answer:

The speed of sound in dry air at 0°C is 331 meters/second

Explanation:

Sound wave is a longitudinal wave. The speed of sound depends on the density and compressibility of the medium. If the density of the medium is less, then it will have faster speed of sound and if the compressibility of the medium is high, it will have slower speed of sound.

It also depends on the temperature. At 20 degrees Celsius, the speed of sound is 343 m/s. At 25 degrees Celsius, its speed is 346 m/s and at 0 degrees Celsius, the speed of sound in dry air is 331.3 m/s

The formula is as follows :

v=(331.3+0.606\times T)\ m/s

T = 0°C

v = 331.3 m/s

or

v = 331 m/s

So, the correct option is (b) "331 meters/second".

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Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

W=Fd

Substitute the value in the above equation.

\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

Thus, the mass of the box is 12.9 kg.

7 0
9 months ago
Question 10 of 10 When light passes through a small opening in a barrier, it bends to spread out on the other side of the openin
suter [353]

Answer:

C. It is the result of diffraction

Explanation:

I just took the test

7 0
2 years ago
Read 2 more answers
Some tropical butterflies can reach speeds of up to 11 m/s. Suppose a butterfly flies at a speed of 6.0m/s while another flying
ladessa [460]

Answer:

24.3 m

Explanation:

Using the equation of motion

S=ut+0.5at^{2} where s is the distance, u is initial velocity, t is time and a is acceleration

Substituting u for 6 m/s, t for 3 s, a for 1.4 m/s2 we obtain

S=6*3+(0.5*1.4*3^{2}=18+6.3=24.3 m


7 0
2 years ago
In a 200-turn automobile alternator, the magnetic flux in each turn is ΦB = 2.50 10-4 cos ωt, where ΦB is in webers, ω is the an
n200080 [17]

Answer:

10.63952sin(209.43951t)

10.63952 V

Explanation:

N_t = Number of turns = 200

\phi_B = Magnetic flux = 2.5\times 10^{-4}cos(\omega t)

\omega_e = Engine angular speed = 1\times 10^{3}\ rpm

Alternator angular speed is given by

N_a=2\times \omega_e\\\Rightarrow N_a=2\times 1\times 10^{3}\\\Rightarrow N_a=2\times 10^{3}\ rpm

\omega=N_a\dfrac{2\pi}{60}\\\Rightarrow \omega=2\times 10^{3}\dfrac{2\pi}{60}\\\Rightarrow \omega=209.43951\ rad/s

Induced emf is given by

\epsilon=-N_t\dfrac{d\phi_B}{dt}\\\Rightarrow \epsilon=-200\dfrac{d}{dt}2.54\times 10^{-4}cos(209.43951 t)\\\Rightarrow \epsilon=200\times 2.54\times 10^{-4}\times 209.43951 sin(209.43951t)\\\Rightarrow \epsilon=10.63952sin(209.43951t)

The function is 10.63952sin(209.43951t)

The induced maximum emf is 10.63952 V

5 0
2 years ago
Please help me! I really don't understand physics. (Please explain how you got the answers also) Thank you!
pickupchik [31]
P = Momentum
M = Mass
t = time
v = velocity (speed)
J= Impulse

1.)
F = 100N
t = 5s

Change in momentum(P) = Ft

I see that this is not one of your equations but your equations do not work with this problem.

P = Ft
P = (100)(5)
P = 500 Ns

2.)
See if you can get this one by yourself from what I did on the last problem.


3.)
P = 25 kgm/s
M = 2.0 kg

So, you are looking for velocity (speed). Look for the equation that has momentum, mass and velocity.

Use the equation P = mv

25 = 2(v)
25 = 2v
25/ 2 = 2v/2
12.5 m/s = v

4.) Do the same for number 4.
P = 12 kgm/s
M = 1kg

P = mv
12 = 1v
12/1 = 1v/1
12 m/s = v

6.)
P = 100 kgm/s
V= 7.0m/s

P= mv
100 = m7.0
100/7 = 7.0m/7
14.28 kg = m

Try page 2 by yourself and message me or comment if you need more help. Just identify the variable you have and what you need to find and put them in an equation that fits. You can do this!!!!


3 0
2 years ago
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