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4 years ago

What is the speed of sound in air when the temperature is 20°C?

Physics

2 answers:

Anarel [89]4 years ago

5 0

The answer to your question is 343 m/s

Minchanka [31]4 years ago

4 0

Answer:

The speed of sound in air when the temperature is 20°C is 343 m/s.

Explanation:

Given that,

Temperature T = 20°C = 20+273=293 K

The speed of sound in air is the distance traveled divided by time.

We know that,

The relation between the speed of sound in air and temperature is defined as:

$v= 331\sqrt{\dfrac{T}{273}}$

Where, T = temperature in Kelvin

v = speed of sound in air

We substitute the value into the formula

$v = 331\sqrt{\dfrac{293}{273}}$

$v = 342.9 = 343\ m/s$

The speed of sound in air is 331 m/s at 0° C, 343 m/s at 20°C and 346 m/s at 25°C.

Hence, the speed of sound in air when the temperature is 20°C is 343 m/s.

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The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of

Aleks04 [339]

Answer:

(a) 1, average velocity = -65.6 m/s

2, average velocity = -64.8 m/s

3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given by;

$y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}$

(1)

$y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s$

(2)

$y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s$

(3)

$y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s$

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

5 0

3 years ago

There are two particles with charges +Q and +q. The electric force applied to one of the charges is 10 N. Now, +Q is replaced by

KIM [24]

Answer:

F = 10 N

Explanation:

Assuming that we can treat to both particles as point charges, the magnitude of the force that one charge exerts on the other, must obey Coulomb's Law.
The expression for this force, applied to the charges +Q and +q, separated by a distance d, is as follows:

$F_{0} = \frac{k*q*Q}{d^{2}}$

If we replace +Q by +4Q, and d by 2*d, the new expression for the magnitude of the force is as follows:

$F_{1} = \frac{k*q*4*Q}{(2*d)^{2} } = \frac{4}{4} *\frac{k*q*Q}{d^{2} } = F_{0} = 10 N$

As it can be seen, the force will be as same as before making changes, so F= 10 N.

6 0

3 years ago

A rubber ball with a mass of 0.145 kg is dropped from rest. From what height (in m) was the ball dropped, if the magnitude of th

Elena-2011 [213]

Answer:

1.55 m

Explanation:

Momentum: This can be defined as the product of mass of a body and it velocity. the S.I unit of momentum is kgm/s.

Mathematically,

Momentum can be represented as,

M = mv................................. Equation 1

Where m = mass of the body, v = velocity of the body, M = momentum.

Making v the subject of the equation,

v = M/m........................................... Equation 2

Given: M = 0.80 kg.m/s, m = 0.145 kg.

Substituting into equation 2,

v = 0.8/0.145

v = 5.52 m/s.

Using the equation of motion,

v² = u² + 2gs ....................... Equation 3.

Where v = final velocity of the rubber ball, u = initial velocity of the rubber ball, s = distance, g = acceleration due to gravity.

Given: v = 5.52 m/s, u = 0 m/s, g = 9.81 m/s².

Substituting into equation 2

5.52² = 0² + 2(9.81)s

30.47 = 19.62s

s = 30.47/19.62

s = 1.55 m.

Thus the ball was dropped from a height of 1.55 m

8 0

4 years ago

Which part of the female reproductive system produces ova?

vovangra [49]

A would be the best answer beacuse
Ovaries - small, oval-shaped glands that are located on either side of the uterus. The ovaries produce eggs (ova - an ovum is one egg, ova means multiple eggs.) The ovaries also produce the main female sex hormones which are released into the bloodstream.

5 0

3 years ago

Read 2 more answers

Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia

earnstyle [38]

Answer:

The acceleration of rocket B is -22.24 m/s²

Explanation:

Two rockets are flying in the same direction and are side by side at the

instant their retrorockets fire

That means they started from the same point at the same time

Rocket A has an initial velocity of 5800 m/s

Rocket B has an initial velocity of 8600 m/s

After a time t both rockets are again side by side, the displacement of

each being zero

That means they are in the same position again → s = 0

The acceleration of rocket A is -15 m/s²

We need to find the acceleration for rocket B

We can find the time from the information of rocket A by using the

rule → s = u t + $\frac{1}{2}$ a t²

where s is the displacement, u is the initial velocity, a is the

acceleration, and t is the time

→ s = 0 , u = 5800 m/s , a = -15 m/s²

Substitute these values in the rule

→ 0 = 5800 t + $\frac{1}{2}$ (-15) t²

→ 0 = 5800 t - 7.5 t²

Add 7.5 t² for both sides

→ 7.5 t² = 5800 t

Divide both sides by 7.5 t

→ t = 773.3 s

The time for the both rocket to have displacement zero is 773.3 s

Now we can find the acceleration of the rocket B by using the same

rule above

→ u = 8600 m/s , t = 773.3 , s = 0

→ 0 = (8600)(773.3) + $\frac{1}{2}$ a (773.3)²

→ 0 = 6650380 + 298996.445 a

Subtract 6650380 from both sides

→ -6650380 = 298996.445 a

Divide both sides by 298996.445

→ a = -22.24 m/s²

<em>The acceleration of rocket B is -22.24 m/s²</em>

6 0

4 years ago