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3 years ago

What is the speed of sound at 33 °C (m/s)? For a frequency of 5 kHz, how large do you expect the wavelength to be (m)?

Physics

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Answer:

-The speed of sound at 33°C is 362.8 m/s.

-The wavelength at a frequency at 5 kHz is 0.07256 m .

Explanation:

let v = 343 m/s be the speed of sound.

let T be the temperature.

then the speed of sound V, at 33°C is given by:

V = v + 0.6×T

= 343 + 0.6×33

= 362.8 m/s

Therefore, the speed of sound at 33°C is 362.8 m/s.

the wavelength at a frequency of f = 5kHz = 5000 Hz is given by:

λ = V/f

= (362.8)/(5000)

= 0.07256 m

Therefore, the wavelength at a frequency at 5 kHz is 0.07256 m .

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A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan

PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d) V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

55 = 0 + 0 + 0.5*(9.81)*t^2

t = sqrt ( 55 * 2 / 9.81 )

t =3.349 s

- Use the second equation of motion in x direction:

x(f) = x(0) + V_x,i*t

150 = 0 + V_x,i*3.349

V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

V_y,f = V_y,i + g*t

V_y,f = 0 + 9.81*3.349

V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

V^2 = V_y,f^2 + V_x,i^2

V = sqrt (32.85^2 + 44.8^2)

V = 55.55 m/s

5 0

3 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

3 years ago

The net effect of each cycle of the sodium-potassium pump is to ____.? ?decrease the number of positively charged ions within th

stealth61 [152]

The correct answer may be A

3 0

3 years ago

Is lateral shift or lateral displacement same ?

xxMikexx [17]

Answer:

When a ray of light passes through a glass slab of a certain thickness, the ray gets displaced or shifted from the original path. This is called lateral shift/displacement.

Explanation:

7 0

3 years ago

You accelerate a 0.43kg football 5m/s2. Calculate the force you applied to the football.

Elza [17]

The force applied on the football is 2.15 Newton.

Given the data in the question;

Mass of football; $m = 0.43kg$

Acceleration; $a = 5m/s^2$

Force applied; $F = \ ?$

To determine the force applied on the football, we use Newton's laws of motion:

$F = m * a$

Where m is the mass of the object and a is the acceleration.

We substitute our given values into the equation

$F = 0.43kg\ *\ 5m/s^2\\\\F = 2.15kg.m/s^2\\\\F = 2.15N$

Therefore, the force applied on the football is 2.15 Newton.

Learn more: brainly.com/question/2388393

7 0

2 years ago