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3 years ago

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A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

drill’s angular acceleration. Answer in units of rad/s 2

1 answer:

mylen [45]3 years ago

4 0

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

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A person wears a hearing aid that uniformly increases the sound level of all audible frequencies of sound by 28.1 dB. The hearin

ladessa [460]

Answer:

I₂ = 2.13 x 10⁻⁸ W/m²

Explanation:

given,

increase in sound level = 28.1 dB

frequency of the sound = 250 Hz

intensity = 3.3 x 10⁻¹¹ W/m²

Intensity delivered = ?

the difference of intensity level is give as

$\beta_2-\beta_1 = 10log(\dfrac{I_2}{I_o}) - 10log(\dfrac{I_1}{I_o})$

$\beta_2-\beta_1 = 10(log(\dfrac{I_2}{I_o}) -log(\dfrac{I_1}{I_o}))$

$\beta_2-\beta_1 = 10(log(\dfrac{I_2}{I_1})$

$28.1= 10(log(\dfrac{I_2}{I_1})$

$log\dfrac{I_2}{I_1}=2.81$

$\dfrac{I_2}{I_1}=10^{2.81}$

I₂ = 645.65 I₁

I₂ = 645.65 x 3.3 x 10⁻¹¹

I₂ = 2.13 x 10⁻⁸ W/m²

4 0

3 years ago

A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

7 0

3 years ago

Give an example of a situation when it would be appropriate to round a number and a situation in which it would not be appropria

Ivan

it would be appropriate to round a number when you are talking about people or animals. Ex. it fits 10 1/2 people. it will most likely fit 9-11 people.

You it is not appropriate to round with money.

Ex. If you owe someone $9.43 you don't just give them $9

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3 years ago

Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st

DedPeter [7]

Answer:

E1 = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q) = +11nC

distance between thin glass rods = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = $\frac{Q}{2\pi e_{0}rL }$

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.01} - \frac{1}{0.03} )$ ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

= $\frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )$

= 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.02} - \frac{1}{0.02} )$

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

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3 years ago

Analogy: atom is to element as ____ is to compound

Masja [62]

Molecule is the answer to what you are looking for

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3 years ago