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2 years ago

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The fundamental frequency of a resonating pipe is 150 Hz, and the next higher resonant frequencies are 300 Hz and 450 Hz. From t

his information, what can you conclude

1 answer:

olasank [31]2 years ago

3 0

Answer:

Explanation:

In case of open organ pipe ,

for fundamental note 2l = wavelength

frequency n = V / Wave length , V is velocity of sound.

f₁ = V / 2I

for first overtone note 2l /2 = wavelength

wavelength = l

frequency = V / wavelength

f₂ = V / l

for second overtone note 2l /3 = wavelength

wavelength = 2l /3

frequency = V / wavelength

f₃ = 3V /2l

f₁ : f₂ : f₃ : : 1 : 2 : 3

In case of closed organ pipe ,

for fundamental note l = 4 x wavelength

frequency n = V / Wave length , V is velocity of sound.

f₁ = V / 4 I

for first overtone note l / = 3 wavelength / 4

wavelength = 4 l / 3

frequency = V / wavelength

f₂ =3 V /4 l

for second overtone note l = 5 wavelength /4

wavelength = 5l /4

frequency = V / wavelength

f₃ = 5 V /4 l

f₁ : f₂ : f₃ : : 1 : 3 : 5.

In case of open organ pipe , frequencies are in any integral ratio.

In case of closed organ pipe , frequencies are only in odd multiple

Since the given ratio are in any integral , the organ pipe must be open organ pipe.

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A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma

Nesterboy [21]

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

= g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x .50 x ( 10+6 )

= 78.4 N

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2 years ago

While it’s impossible to design a perpetual motion machine, that is, a machine that keeps moving forever, come up with ways to k

MissTica

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1 year ago

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Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he sw

Natasha_Volkova [10]

Answer:

Explanation:

Given

average speed of Phelps $v_{avg}=2\ m/s$

total distance $d=200\ m$

he swims first 100 m at an average speed if $1.8 m/s$

so time taken is $t_1=\frac{100}{1.8}=55.55\ s$

suppose $t_2$ is the time taken to swim remaining half

average velocity is $v_{avg}=\frac{displacement}{total\ time}$

$v_{avg}=\frac{100+100}{55.55+t_2}$

$t_2+55.55=\frac{200}{2}=100$

$t_2=44.45\ s$

so velocity in the second half is

$v_2=\frac{100}{45.45}$

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1 year ago

What is the magnitude of the relative angle φ

melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

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1 year ago

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

We need to calculate the velocity of the ball

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

Using conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

Put the value into the formula

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

$d =vt$

Where. d = distance

t = time

v = velocity

Put the value into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0

2 years ago