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daser333 [38]
3 years ago
9

The speed of sound is 346 m/s. If a sound wave travels at a frequency of 55 Hz, what would its wavelength be?​

Physics
1 answer:
Korvikt [17]3 years ago
3 0

Answer:

6.29 meters.

Explanation:

, where v is the speed of wave and f is the frequency of wave.

We are given that ,

The speed of sound is 346 m/s.

i..e v=346 m/s

A sound wave travels at a frequency of 55 H.

i..e f=55

the wavelength would be 6.29 meters.

This is based on another brainly answer

Link: brainly.com/question/12538018

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ListenA bicycle and its rider have a combined mass of 80. kilograms and a speed of 6.0 meters per second. What is the magnitude
Setler [38]

Answer:

a) 1.2\times 10^2\ N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-6}{4}\\\Rightarrow a=-1.5\ m/s^2

The acceleration of the bicycle and rider is -1.5 m/s²

Force

F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N

The magnitude of the average force needed to bring the bicycle and its rider to a stop is 1.2\times 10^2\ N

3 0
3 years ago
Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.
Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

F=\frac{k.q.Q}{r^{2}}

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}

F_{12}=536.02.10^{-3} N

Force between 2 and 3:

F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}

F_{23}=2711.63.10^{-3} N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

F_{T}=2711.63.10^{-3}-536.02.10^{-3}

F_{T}=2175.61.10^{-3} N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, F_{13}=F_{23}. Suppose distance from 1 to 3 is x, then from 2 to 3 is x-0.301

Calculating:

\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}

\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}

\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}

x^{2}=0.67x^{2}-0.40x+0.061

0.33x^{2}+0.40x-0.061=0

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0
3 years ago
Help and show work thanks
Zepler [3.9K]
Well its C, cant show you the work its in my head sorry.
6 0
3 years ago
Typical Pressurized Water Reactors can produce 1100 to 1500
lozanna [386]

Answer: about 1,100,000,000 to 1,500,000,000 Joules/second

Explanation:

1 MW (megawatt) = 1,000,000.00 J/s (joules per second)

1100(1,000,000) = 1,100,000,000

1500(1,000,000) = 1,500,000,000

3 0
2 years ago
A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c
satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

First course : a_{x} = 0.75 m/s², d_{x} = 20 m, u_{x} = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

v_{x}² = (10)² + (2 × 0.75 × 20)

v_{x}² = 100 + 30

v_{x}² = 130

v_{x} = √130

v_{x} = 11.4 m/s

for the Second Course:

u_{y} =  11.4 m/s,  a_{y} = -1.15 m/s²,  v_{y} = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×  d_{y} )

0 = 129.96 - 2.3d_{y}

2.3d_{y}  = 129.96

d_{y} = 129.96 / 2.3

d_{y} = 56.5 m

so;

|d| = √( d_{x}² + d_{y}² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0
2 years ago
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