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3 years ago

How fast is a wave traveling if it has a wavelength of 7 meters and a frequency of 11 Hz?

Physics

1 answer:

pashok25 [27]3 years ago

7 0

Answer:

$\huge{ \boxed{ \bold{ \sf{77 \: m/s}}}}$

☯ Question :

How fast is a wave travelling if it has a wavelength of 7 meters and a frequency of 11 Hz?

☯ $\underbrace{ \sf{Required \: Answer : }}$

☥ Given :

Wavelength ( λ ) = 7 meters
Frequency ( f ) = 11 Hz

☥ To find :

Speed of sound ( v ) = ?

☄ We know ,

$\boxed{ \sf{v = f \times λ}}$

where ,

v = speed of sound
f = frequency
λ = wavelength

Now, substitute the values and solve for v.

➺ $\sf{v = 11 \times 7}$

➺ $\boxed{ \sf{v = 77 \: m/s}}$

-------------------------------------------------------------------

✑ Additional Info :

Frequency : The number of complete vibrations made by a particle of a body in one second is called it's frequency. It is denoted by the letter f . The SI unit of frequency is hertz ( Hz ).

Wavelength : The distance between two consecutive compressions or rarefactions of a sound wave is called wavelength of that wave. It is denoted by λ ( lambda ) and it's SI unit is m.

Speed of a sound wave : The distance covered by a sound wave in one second is called speed of sound wave. It depends on the product of wavelength and frequency of the wave.

Hope I helped!

Have a wonderful time! ツ

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3 years ago

Solve for the length of the inclined plane if the angle equals 19.45 degrees.

mel-nik [20]

The length of the inclined plane is approximately 12 ft

The situation forms a right angle triangle.

<h3>Right triangle</h3>

Right triangle have one of its angle as 90 degrees.

Therefore,

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height = 4 ft

angle(∅) = 19.45°

sin 19.45 = 4 / h

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2 years ago

A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.

jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

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2 years ago

Read 2 more answers

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh

irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

$K=\frac{1}{2}mv^2$

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

$U=mgh$

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

$a=\frac{v-u}{t}$

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

$u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s$

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

$\Delta v = 0 - (+9.8 m/s)=-9.8 m/s$

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

$a=\frac{v-u}{t}$

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

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and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - 0=-9.8 m/s$

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

$\Delta v = v -u$

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s$

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

$mg = ma$

which means that the acceleration is

$a= g = -9.8 m/s^2$

and the negative sign means it points downward.

7 0

3 years ago

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LekaFEV [45]

Answer:

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Explanation:

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Subtract

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The measured pressure is in mm Hg to take this unit to units of pressure must be multiplied by the density of mercury and the acceleration of gravity

P₁- P₂ = rho_Hg g (h₁-h₂) _Hg

rho_Hg g (h₁-h₂) _Hg = roh_air g (h₁ –h₂)

(h₁ –h₂) = rho_Hg / rho_air (h₁-h₂) _ Hg

Let's calculate

(h₁-h₂) = 13600 / 1.18 (695-675)

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Let's reduce to meter

(h₁-h₂) = 2.30 10⁵ mm (1 m / 10³ mm)

(h₁-h₂) = 2.30 10² m

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3 years ago