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3 years ago

A sound wave has a frequency of 628 Hz in air

Physics

1 answer:

dedylja [7]3 years ago

8 0

Answer:

Answer is 257°C

Explanation:

Given

frequency = 628 Hz

wavelength = 0.49 m

speed of sound at 0°C = 325 m/s

Find

1.speed of sound at t °C=?

2.temperature = ?

Formula

speed of sound at t °C = speed of sound at 0°C [1+t/273]

Solution

1. speed of sound at t °C = f*∧

= 628*0.49

=307.7m/s

2.speed of sound at t °C = speed of sound at 0°C [1+t/273]

307.7 m/s = 325 m/s [1+ t/ 273]

307.7/325 = 1+t/273

0.94 = 1+t/273

multiplying 273 on both sides

(0.94)(273) = 1+t/273 (273)

258 = 1+ t

258 - 1 = t

257°C = t

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The mass of a basketball is three times greater than the mass of a softball . Compare the momentum’s of a softball and a basketb

Lemur [1.5K]

They are moving the softball of basketball

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3 years ago

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?

Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

Q = 0.456 x 33.5 x 10^(4) = 152760 J

the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

6 0

4 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

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3 years ago

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liberstina [14]

Answer:50kg

Explanation:

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3 years ago

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harina [27]

Answer:

False

Explanation:

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4 0

4 years ago

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