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3 years ago

A ball is thrown straight up. For which situation are both the instantaneous velocity and the acceleration zero?

Physics

1 answer:

leva [86]3 years ago

6 0

The correct option is b i.e none is correct.

The instantaneous velocity would be zero at the top but the acceleration would not be zero at that point. This is because the acceleration is the rate of change of velocity. since, the velocity is not constant, hence the acceleration would not be zero. The correct answer would be option b: None is correct.

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A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the east

Marrrta [24]

Answer:

ax = 6.43m/s²

Explanation:

The acceleration is the time derivative of the velocity function ax = dvx(t)/dt

We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.

So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.

7 0

4 years ago

Imagine a ringing bell set inside a sealed glass jar. Once all the air is removed and a vacuum is crated, the ringing sound is n

kenny6666 [7]

working...

Sound wave needs medium to travel

as energy which travels in this wave is because of transfer from one particle to another particle

If there is no medium then energy can not be transferred and sound wave will not travel

so in vacuum we can not listen sound

similarly here air is removed it means there is no medium inside the jar to travel the sound and hence we can not hear it

Option B is correct

Without air, the sound waves cannot travel to the ear.

5 0

4 years ago

(6) Module 4 Study Guide

velikii [3]

We have the meats Arby’s we beat them kids

3 0

3 years ago