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nekit [7.7K]
3 years ago
11

Avg velocity definition | Problem Average velocity vectors…

Physics
1 answer:
Aneli [31]3 years ago
3 0

Answer:

point in the same direction as the change in position.

require knowing the initial and final position, along with the elapsed time.

Explanation:

As we know that average velocity is defined as the ratio of change in position and the time interval of the object

It is given as

v_{avg} = \frac{\Delta x}{\Delta t}

so we will have

\Delta x = displacement

\Delta t = time interval

the direction is this average velocity is always along the direction of the displacement

so correct answer will be

point in the same direction as the change in position.

require knowing the initial and final position, along with the elapsed time.

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I'll make you brainliest! EaSyQuEsTiOn pleaze ASAP hurry! Make you brainliest Promise!!!
inysia [295]

Answer:

It hit Earth like a really long time ago, so many rocks and soil should have been piled on top of the crater because of wind, rain, etc.

Also, it is really deep and really old so when it is old, it gets less visible to the naked eye.

8 0
3 years ago
The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit
sveticcg [70]

Answer:

a)   I = 3.63 W / m² , b)   I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

         I = P / A = ½ ρ v (2π f s_{max})²

         I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

        cte = (½ ρ v 4π² s_{max}²)

         I₀ = cte s² f₀²

        I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

         I = constant (2.20 10³) 2

         I = cte 4.84 10⁶

let's find the relationship of the two quantities

        I / Io = 4.84

        I = 4.84 Io

        I = 4.84 0.750

        I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

        I = cte (f s)²

        I = constant (0.250 10³ 4)²

 

        I = cte 1 10⁶

         

the relationship

        I / Io = 1

        I = Io

        I = 0.750 W / m²

6 0
2 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values
Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level  β  = 10 log(I/I_{0})        ......................a

put here value (I/I_{0}) = 10^−12 W/m² and  β = 115

115  = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r²    ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m    = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0
3 years ago
The angular speed of digital video discs (DVDs) varies with whether the inner or outer part of the disc is being read. (CDs func
Kitty [74]

Answer:

α = 0.0135 rad/s²

Explanation:

given,

t = 133 min = 133 x 60 = 7980 s

angular speed varies from 570 rpm to 1600 rpm

now,

570 rpm = 570 \times \dfrac{2\pi}{60}

              = 59.69 rad/s

1600 rpm =  = 570 \times \dfrac{2\pi}{60}

              = 167.6 rad/s

using equation of rotational motion

ωf = ωi + αt    

167.6 = 59.7 + α x 7980

α  x 7980 = 107.9

α = 0.0135 rad/s²

8 0
3 years ago
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