All categories

2 years ago

(42)What is the speed of sound in air at a temperature of 15°C?

Physics

1 answer:

VLD [36.1K]2 years ago

4 0

Answer:

340 m/s

At the temperature of 15 degrees Celsius, the velocity of sound is about 340 m/s in air.

You might be interested in

Plzzz answer this question correctly

tekilochka [14]

Answer:

changing the direction in which a force is exerted

8 0

3 years ago

Rate at which wave travels or propagates

Softa [21]

Answer:

v =fλ v = f λ where v

Explanation:

4 0

3 years ago

A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13

tia_tia [17]

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

$F = qvB$

here we have

$B = \frac{F}{qv}$

here we know that

$F = 4.8 \times 10^{-13} N$

$q = 1.6 \times 10^{-19} C$

$v = 4 \times 10^6 m/s$

now from above equation we have

$B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}$

$B = 0.75 T$

8 0

3 years ago

Please help i'm going to throw up from stress

Eddi Din [679]

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

$t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})}$ where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

$16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)}$ and

$16.0=\frac{83640x+2928}{4182x+292.8}$ and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

4 0

3 years ago

3 A 100 g steel ball falls from a height of 1.8 m on to a metal plate and rebounds to a height of 1.25 m.

BigorU [14]

Given values:

Mass of the steel ball, m = 100 g = 0.1 kg

Height of the steel ball, h1 = 1.8 m

Rebound height, h2 = 1.25 m

a. PE= mgh

0.1 x 9.8 x 1.8 =

1.764 Joules

b. KE = PE ->

1.764 Joules

c. KE= 1/2 mv square

so v = square root 2ke/m

square root 2 x 1.764/ 0.1

= 5.93 m/s

d. KE=PE=mgh square

0.1 x 9.8 x 1.21 =

1.186 joules

velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s

6 0

3 years ago

Read 2 more answers