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3 years ago

Which of these objects would be most likely to create this sound wave?

Physics

1 answer:

saw5 [17]3 years ago

3 0

The image is missing, so i have attached it.

Answer:

Correct answer is Option C - Ticking clock

Explanation:

Sound waves are formed when a vibrating object makes the encompassing medium to vibrate. Where the medium is the material in which the wave goes through. Sound waves comprise of zones of high and low weight called compressions and rarefactions, separately.

Now, in this question, the image given is one of compression and rarefactions. Looking at the options given, the correct option that produces such a wave is Ticking clock because Sound energy travels through air in waves and when an alarm clock rings, nearby air molecules vibrate. The air molecules jiggle about and cause neighbouring molecules to vibrate as well, thereby starting a wave of energy that travels out from the clock. The wave of sound travels in a pattern of compressions (where the air molecules are squeezed together) and rarefactions (where the air molecules are stretched apart).

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show mathematically why an 80,000 pound ( 36,000 kg ) big rig taveling 2 mph (0.89 m/s) has the SAME MOMENTUM as a 4,000 pound (

bekas [8.4K]

The momentum value for both Big rig and sport utility vehicle are the same as 16000 pound mph

<u>Explanation:</u>

Given data are as follows

For Big rig, Mass = 80,000 pound (36,000 kg)

velocity = 2 mph (0.89 m/s)

where mph is meter per hour

For sport utility vehicle, Mass = 40,000 pound (1800 kg)

velocity = 40 mph (18 m/s)

The formula to find the Momentum of an object is

Momentum = Mass × Velocity (Kilogram meter per second)

Momentum for Big rig = 80,000 × 2 (pound mph)

= 1,60,000 pound mph

Momentum for sport utility vehicle = 4000 × 40 (pound mph)

= 1,60,000 pound mph

Hence it is mathematically proved that

The momentum of big rig = The momentum of sport utility vehicle

1,60,000 pound mph = 1,60,000 pound mph

3 0

4 years ago

Convert 9.2 km to m

Nataly_w [17]

Answer:

9,200 m

Explanation:

1 kilometer = 1000 meters so if there is 9 kilometers then it is 9000 meters, then .2 kilometer is equal to 200 meters. Add the numbers together to get 9,200 meters

5 0

4 years ago

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

5 0

3 years ago

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then t

lapo4ka [179]

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_{atm}$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}$

Where:

$P_{1}$ , $P_{2}$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the tank and ground level, measured in meters.

Given that $P_{1} = P_{2} = P_{atm}$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 0\,\frac{m}{s}$ , $z_{1} = 6.9\,m$ and $z_{2} = 4.9\,m$ , the expression is reduced to this: