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Aleksandr-060686 [28]
2 years ago
10

The speed of sound in dry air at 20 °C is 343.5 m s-1, and the frequency of the sound from the note C# above middle C on the pia

no is 277.2 s-1(according to the American standard pitch scale). Calculate the wavelength of the sound and the time it will take to travel 49.2 m across a concert hall. Wavelength = (answer) m
Physics
1 answer:
Nastasia [14]2 years ago
4 0

Answer:

1.23917 m

0.14323 s

Explanation:

v = Speed of sound in dry air at 20 °C = 343.5 m/s

f = Frequency of note C# = 277.2 /s = 277.2 Hz

λ = Wavelength

v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343.5}{277.2}\\\Rightarrow \lambda=1.23917\ m

Wavelength = 1.23917 m

Distance the wave needs to travel is 49.2 m

Time = Distance / Speed

\text{Time}=\frac{49.2}{343.5}=0.14323\ s

Time taken for the sound to travel across the concert hall is 0.14323 s

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astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

8 0
2 years ago
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45
Svet_ta [14]

Answer:

Kindly check explanation

Explanation:

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.

Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

5 0
2 years ago
The atomic mass of sodium is 0.0230 kg/mole. How many moles are in 1.59 kg of sodium?
Evgesh-ka [11]

Answer:

0.0144..

Explanation:

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6 0
1 year ago
Read 2 more answers
Can someone help with my physics homework? please
Murrr4er [49]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

W = F\cdot \Delta s (1)

Where:

F - Force applied by the cart, measured in newtons.

\Delta s - Distance travelled by the car, measured in meters.

W - Work done on the cart, measured in joules.

If we know that F = 888\,N and \Delta s = 22\,m, then the work done on the cart is:

W =(888\,N)\cdot (22\,m)

W = 19536\,J

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that F = 1111\,N and \Delta s = 22\,m , then the work on the cart is:

W = (1111\,N)\cdot (22\,m)

W = 24442\,N

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

3 0
2 years ago
A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest
IRINA_888 [86]

Answer:Average speed is greater than average velocity.

Explanation  :

Given

student walks slowly along a straight line for a while ,then stops to rest a while, and finally runs quickly back to her initial position

Let x be the length of track and the whole process takes t time

For average speed =\frac{distance\ traveled}{time\ taken}

Average speed=\frac{2x}{t}

For average velocity =\frac{Displacement}{time\ taken}

Since displacement is zero as she returns to its initial position.

Average velocity=0

Therefore Average speed is greater than average velocity.

5 0
2 years ago
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