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r-ruslan [8.4K]
4 years ago
8

Explain why there is a delay between the arrival of the sound transmitted through the steel pipe and the sound transmitted by th

e air in the pipe.​
Physics
2 answers:
Tamiku [17]4 years ago
8 0

The speed of sound in steel is about 17 times as fast as it's speed in air.

Naturally, when a sound starts out at one end of a pipe, it reaches the other end thru the steel before it reaches the other end thru the air in the pipe.

The longer the pipe is, the longer will be the delay between the arrival of the two sounds at the other end.

Nataliya [291]4 years ago
7 0

Answer: RATE ME AND MAKE ME BRAININESS AND THANK ME

Explanation:Sound waves are pressure waves that travel through Earth's crust, water bodies, and atmosphere. Natural sound frequencies specify the frequency attributes of sound waves that will efficiently induce vibration in a body (e.g., the tympanic membrane of the ear) or that naturally result from the vibration of that body.

Sound waves can potentiate or cancel in accord with the principle of superposition and whether they are in phase or out of phase with each other. Waves of all forms can undergo constructive or destructive interference. Sound waves also exhibit Doppler shifts—an apparent change in frequency due to relative motion between the source of sound emission and the receiving point. When sound waves move toward an observer the Doppler effect shifts observed frequencies higher. When sound waves move away from an observer the Doppler effect shifted observed frequencies lower. The Doppler effect is commonly and easily observed in the passage of planes, trains, and automobiles.

The speed of propagation of a sound wave is dependent upon the density of the medium of transmission. Weather conditions (e.g., temperature , pressure, humidity , etc.) and certain geophysical and topographical features (e.g., mountains or hills) can obstruct sound transmission. The alteration of sound waves by commonly encountered meteorological conditions is generally negligible except when the sound waves propagate over long distances or emanate from a high frequency source. In the extreme cases, atmospheric conditions can bend or alter sound wave transmission.

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Your friend (68 kg) is wearing frictionless roller skates and standing still. You throw at her a 3.6 kg pumpkin with a velocity
Fudgin [204]

Answer:

Approximately 0.48\; \rm m\cdot s^{-1}.

Explanation:

Momentum would be conserved since there's no friction on this friend, and all other forces on her are balanced. Therefore:

\begin{aligned}& \text{Resultant momentum of (friend and pumpkin)} \\ =\; & \text{Initial momentum of (friend)} \\ & + \text{Initial momentum of (pumpkin)}\end{aligned}.

Momentum p the product of mass m and velocity v. That is: p = m \, v.

The initial momentum of this friend is 0\; \rm kg \cdot m \cdot s^{-1} since she was initially not moving (an initial velocity of 0\; \rm m\cdot s^{-1}.)

The initial momentum of the pumpkin would be:

\begin{aligned}p &= m \, v \\ &= 3.6 \; \rm kg \times 9.5\; \rm m\cdot s^{-1} \\ &= 34.2\; \rm kg \cdot m \cdot s^{-1}\end{aligned}.

Therefore:

\begin{aligned}& \text{Resultant momentum of (friend and pumpkin)} \\ =\; & \text{Initial momentum of (friend)} \\ & + \text{Initial momentum of (pumpkin)} \\ =\; &0\; {\rm kg \cdot m \cdot s^{-1}} + 34.2\; {\rm kg \cdot m \cdot s^{-1}} \\ =\; & 34.2\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}.

Rearrange the equation p = m \, v to find an expression for velocity v given momentum and mass:

\displaystyle v = \frac{p}{m}.

Note that the "final momentum of friend and pumpkin" in the previous equation refers to the resultant velocity of the friend with the pumpkin in her hand. Thus, it would necessary to use the combined mass of the friend and the pumpkin (68\; {\rm kg} + 3.6 \; {\rm kg}) when calculating the resultant velocity:

\begin{aligned}& \text{Resultant velocity of (friend and pumpkin)} \\ =\; & \frac{\text{Resultant momentum of (friend and pumpkin)}}{\text{Mass of (friend and pumpkin)}} \\ =\; & \frac{34.2\; {\rm kg \cdot m \cdot s^{-1}}}{68\; {\rm kg} + 3.6\; {\rm kg}} \\ \approx \; & 0.48\; \rm m \cdot s^{-1}\end{aligned}.

6 0
3 years ago
A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder a
jonny [76]

Answer:

V = 2.64 10⁻⁴ m³,    T = 21.85 N ,    T = 19.26 N

Explanation:

To calculate the cylinder volume we use the density equation

     ρ = m / V

    V = m / ρ

    ρ = 8.44 g / cm³ (1 kg / 1000g) (10² cm / 1m)³ = 8.44 10³ kg / m³

    V = 2.23 / 8.44 10³

    V = 2.64 10⁻⁴ m³

We calculate the tension with Newton's second law

In the air

    T-W = 0

    T = mg

    T = 2.23 9.8

    T = 21.85 N

In water

In this case we have the push up

   T + B -W = 0

   T = W -B

The push formula is

   B = \rho_{water}  g V

   T = m g - \rho_{water}  g V

   T = 2.23 9.8 - 1.00 103 9.8 2.64 10-4

   T = 21.85 - 2.5872

   T = 19.26 N

3 0
4 years ago
Tigers and goldfish are not related
Lesechka [4]
But they are ‍♀️‍♀️ because ima idk they just are search it up sis
8 0
4 years ago
I need help
frutty [35]

Newton's second law allows finding the answer for the force that attracts the body is 127.4N

Newton's second law indicates that the force that is the interaction between two bodies is directly proportional to the product of the mass and the acceleration.

            F = m a

Where the bold letters indicate vectors, F is the force, m the mass, and the acceleration of the body.

When a body is close to the Earth there is an interaction between the body and the planet, we call this interaction weight, it is given by the relationship

           W = m g

Where W is the force called weight, m ​​the mass of the body and g the acceleration of the body, which in this case is called the gravity acceleration   (g = 9.8 m / s²)

They indicate that the mass of the body is m = 13 kg, let's calculate the weight

           W = 13  9.8

           W = 127.4 N

In conclusion using Newton's second law we can find the answer for the force that attracts the body is 127.4N

Learn more here:

brainly.com/question/19226427

3 0
2 years ago
A 98-kg fullback is running along at 8.6 m / s when a 76-kg defensive back running in the same direction at 9.8 m / s jumps on h
algol13
The total momentum before and after the collision must be conserved.

The total momentum before the collision is:
p_i = m_1 v_1 + m_2 v_2
where m1 and m2 are the masses of the two players, and v_1 and v_2 their initial velocities. Both are considered with positive sign, because the two players are running toward the same direction.

The final momentum is instead
p_f = (m_1+m_2)v_f
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.

By requiring that the momentum is conserved
p_i=p_f
we  can calculate vf, the post-collision speed:
m_1 v_1 + m_2 v_2 = (m_1+m_2)v_f
v_f =  \frac{m_1 v_1 + m_2 v_2}{m_1 +m_2}= \frac{(98 kg)(8.6 m/s)+(76 kg)(9.8m/s)}{98 kg+76 kg}=9.1 m/s
and the direction is the same as the direction of the players before the collision.
6 0
3 years ago
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