PHYSICS COLLEGE

A vibrating tuning fork makes 500 vibrations in one second. What is the wavelength of the sound produced if the air temperature is 20°C? Group of answer choices0.500 m
0.686 m
0.343 m
1.46 m
1.87 m

Answers

Answer 1
Answer:

Answer:

wavelength will be 0.686 m

So option (b) is correct

Explanation:

We have given vibration in one second that is frequency = 500 Hz

We know that velocity of sound is 343 m/sec

We have to find the wavelength

We know that velocity is given by v=\lambda f

So wavelength \lambda =(v)/(f)=(343)/(500)=0.686m

So wavelength will be 0.686 m

So option (b) is correct

Related Questions

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.
At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.
Mudflows composed of soil, volcanic debris, and water can occur as the result of an explosive volcanic eruption. What are these mudflows called?
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.50 m/s and observes that it takes 1.2 s to reach the water. How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Round your answer to the nearest whole number.
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 162 cm , but its circumference is decreasing at a constant rate of 14.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.1 m/s2 while Kathy maintains an acceleration of 4.99 m/s. 2 (a) Find the time at which Kathy overtakes Stan. s from the time Kathy started driving (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him. Kathy m/s Stan m/s

Answers

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

(c) vf₁ = 15.097 m/s : Stan's final speed

    vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement   

t₁ :  Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁:  Stan acceleration

d₂: car displacement   

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂:  Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ =  1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1=equation 2

1.55*t₁²  =  2.495* t₂²  , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² =2.495* t₂²

1.55* (t₂² +2t₂+1) =2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0  Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant  Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁    t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂  

vf₂ =0+4.99* 3.87

vf₂ = 19.31m/s : Kathy's final speed

A ball connected to a 1.1 m string and is swing in circular fashion. It’s tangential velocity is 15 m/s. What is its centripetal acceleration?

Answers

Answer:

ac = 204 [m/s²]

Explanation:

To solve this problem we must use the following equation that relates the tangential velocity to the radius of rotation.

ac = v²/r

where:

v = tangential velocity = 15 [m/s]

r = radius = 1.1 [m]

Now replacing we have:

ac = (15)²/1.1

ac = 204 [m/s²]

A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.30 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 1.20 ?

Answers

Answer:

Explanation:

Area of crossection, A = 7.80 cm²

Initial magnetic field, B = 0.5 T

Final magnetic field, B' = 3.3 T

Time, t = 1 s

resistance of the coil, R = 1.2 ohm

The induced emf is given by

e=(d\phi)/(dt)=A(B' - B)/(t)

where, Ф is the rate of change of magnetic flux.

e = 7.80 x 10^-4 x (3.3 - 0.5) / 1

e = 2.184 mV

i = e/R

i = 2.184/1.2

i = 1.82 mA

A plank 2.00 cm thick and 13.0 cm wide is firmly attached to the railing of a ship by clamps so that the rest of the board extends 2.00 m horizontally over the sea below. A man of mass 68.4 kg is forced to stand on the very end. If the end of the board drops by 5.20 cm because of the man's weight, find the shear modulus of the wood.

Answers

Answer:

9.93 MPa

Explanation:

Given:

 - mass of the man = 68.4 kg

 - Deflection dx = 5.2 cm

 - thickness of plank t = 2.0 cm

 - width of plank w = 13.0 cm

 - Length subtended L = 2.0 m

Find:

Shear Modulus of Elasticity S :

                 S = shear stress / shear strain

                        Shear stress = F / A

                        Shear stress = 68.4*9.81 / 0.02*0.13

                        Shear stress = 258078.4615 Pa

                        Shear strain = dx / L

                        Shear Strain = 0.052 / 2

                        Shear Strain = 0.026  

Hence,

                  S = 258078.4615 / 0.026

                  S = 9.93 MPa

In a super-heater (A) pressure rises, temperature drops (B) pressure rises, temperature remains constant (C) pressure remains constant and temperature rises (D) both pressure and temperature remains constant

Answers

Answer:

i believe that it is d

Explanation:

Final answer:

In a super heater, the temperature of the steam rises while the pressure remains constant. This process helps to remove the last traces of moisture from the saturated steam.

Explanation:

In a super heater, the conclusion is that option (C) pressure remains constant and temperature rises is the correct choice. A super heater is a device used in a steam power plant to increase the temperature of the steam, above its saturation temperature. The function of the super heater is to remove the last traces of moisture (1 to 2%) from the saturated steam and to increase its temperature above the saturation temperature. The pressure, however, remains constant during this process because the super heater operates at the same pressure as the boiler.

Learn more about Super heater here:

brainly.com/question/32665042

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. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force

Answers

Answer:

elative magnitude of the two forces is the same and they are applied in a constant direction.

Explanation:

Newton's second law states that the sum of the forces is equal to the mass times the acceleration  

              ∑ F = m a

in this case there are two forces on the x axis

             F_applied - fr = 0

since they indicate that the velocity is constant, consequently

             F_applied = fr

the relative magnitude of the two forces is the same and they are applied in a constant direction.
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