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crimeas [40]
3 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]3 years ago
8 0

Answer:

1,450 meters per second

Explanation:

amm18123 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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What is an effect of an object getting wet?
djverab [1.8K]

Answer:

once light hits a wet shirt, that water layer causes less of the blue shirt's blue wavelengths of light to be reflected toward your eyes and more of the blue light to be refracted, or bounce away from you, back into the fabric.

Explanation:

7 0
3 years ago
If τ=r×F then F.τ is equal
Mademuasel [1]
Yes that is correct.
4 0
3 years ago
Four charges are on the four corners of a square. Q1 = +5μC, Q2 = -10μC, Q3 = +5μC, Q4 = -10μC. The side length of the square is
Marat540 [252]

Answer:

Explanation:

Electric field due to a point charge Q at a point at distance d is given by the relation

E = \frac{K\times Q}{d^2}

Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.

The charges are situated on the corners of a square in such a way that

equal charges of Q1 and Q3 are situated on the diametrically  opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two  charges will be zero.  

On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero

Overall, net field due to all the four charges will be zero

3 0
3 years ago
A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on
kari74 [83]

Answer:

9.16rad/s^2

Explanation:

We are given that

Mass,m_1=3.3 kg

Radius,r=0.8 m

m_2=4.9 kg

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

4.9g-T=4.9a

Tr=I\alpha

Where

\alpha=\frac{a}{r}

Tr=\frac{1}{2}m_1ra

T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a

Substitute the value

4.9g-\frac{1}{2}(3.3a)=4.9a

4.9\times 9.8=4.9a+\frac{3.3a}{2}

Where g=9.8 m/s^2

48.02=a(4.9+1.65)=6.55a

a=\frac{48.02}{6.55}=7.33m/s^2

Angular acceleration,\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2

7 0
3 years ago
What is the mass of a baseball thrown at 25 m/s resulting in a momentum of 10 kg·m/s?
Tresset [83]

Answer: m = 0.4 kg

Explanation: Momentum is the product of mass and velocity also expressed in this equation:

p = m x v

Derive to find m

m = p / v

= 10 kg•m/s / 25 m/s

= 0.4 kg

8 0
3 years ago
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