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2 years ago

"Maximum constructive interference occurs at the

Physics

1 answer:

damaskus [11]2 years ago

7 0

The diagram you posted above represents a standing wave on a string. The maximum constructive interference occurs at the antinodes B and D. The correct answer choice is number 3, antinodes B & D. Be careful when choosing as the above choices are not in correct order. Instead of 1,2,3,4, the choices above are 1,3,2, and 4.

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The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. the heat increased the _

uranmaximum [27]

<h2>Apartment Explosion Reported </h2>

The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. The heat increased the temperature of the air in the room, which means an increase in the air's molecular kinetic energy.

When heat is provided then temperature increases and the molecules of substances move rapidly by increase of kinetic energy (K.E) temperature increases. It is understood that heat increases temperature.

6 0

2 years ago

Read 2 more answers

A 10.0 cm3 sample of copper has a mass of 89.6

Romashka-Z-Leto [24]

Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*

4 0

2 years ago

You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac

Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}$

Stop sight distance= 95.07 m

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}$

Stop sight distance= 81.88 m

5 0

2 years ago

At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti

dalvyx [7]

Answer:

Part(a): The angular acceleration is $5.63~rad~s^{-2}$ .

Part(b): The angular displacement is $2629~rad$ .

Explanation:

Part(a):

If $\omega_{1},~\omega_{2}~and~\alpha$ be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

$\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)$

where ' $t$ ' is the time taken by the centrifuge to increase its angular speed.

Given, $\omega_{i} = 250~rad~s^{-1}$ , $\omega_{f} = 750~rad~s^{-1}$ and $t = 9.5~s$ . From equation ( $I$ ), the angular acceleration is given by

$\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}$

Part(b):

Also the angular displacement ( $\Delta \theta$ ) can be written as

$&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad$

8 0

2 years ago

If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee

Alexus [3.1K]

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

7 0

2 years ago