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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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Your outlets at home are rated at 120 V, i.e. the two prongs have on average a potential difference of 120V. If you transfer 2.7
hodyreva [135]

Answer:

E =230.4 MJ

Explanation:

As 1 mole of electron =  6X 10^23 particles.

charge of an electron is 1.6 X 10 ^-19 C

Finding Charge:

(6X10^23 ) (2.7)(1.6X10^-19 C)

i.e. 192 K C

now  to find the energy released from electrons

V=E/q

E=V X q

i.e E = 120 V X 192 K C

E =230.4 MJ

4 0
3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
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PolarNik [594]

Answer:

the radius of the earth in himalayan region is greater than terai reagion. therefore, the value of 'g' at the poles is greater than the value of g at the equator. 12

Explanation:

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6 0
3 years ago
The energy an object acquires when it is exposed to a force is called _____ energy
r-ruslan [8.4K]
I'm pretty sure the energy an object acquires when exposed to a force is known was potential energy. 
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3 years ago
Is telekinesis real??? I really wanna start learning it!
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if you only have to control your chakra and know how to get all your vibes to pass it to objects and it takes time to practice

4 0
3 years ago
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