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Aleksandr [31]
3 years ago
8

What is the speed of sound at the atmospheric temperature of 30°C?

Physics
2 answers:
Rina8888 [55]3 years ago
5 0

Answer:

Option C is the correct answer.

Explanation:

The speed of sound in air is given approximately by  

v=331.4+0.6T_c

Here T_c is the temperature in degree Celsius.

Temperature T_c=30^0C

Substituting we will get

     v = 331.4 + 0.6 x 30 = 349.4 m/s

Option C is the correct answer.

12345 [234]3 years ago
4 0

The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.


First, we determine how long the parcel will fall using:


s = ut + 1/2 at²


where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.


5.5 = (0)(t) + 1/2 (9.81)(t)²

t = 1.06 seconds



A

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Astronaut Rob leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. Rob's destination is a star sys
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Explanation:

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Which of these statements is true about endothermic reactions, but not about exothermic reactions?
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4 0
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6 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18


400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
3 years ago
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