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zloy xaker [14]

3 years ago

7

A stone is thrown, run a velocity of 15m/s is projected of an elevation of 30° to the horizontal calculate the time rate of flig

ht

1 answer:

frozen [14]3 years ago

3 0

Answer:

1.53 seconds

Explanation:

Applying,

T = 2usin∅/g................ Equation 1

Where, T = time of flight, u = initial velocity, ∅ = angle of projectile to the horizontal, g = acceleration due to gravity

From the question,

Given: u = 15 m/s, ∅ = 30°

Constant: g = 9.8 m/s²

Substitute these values in equation 1

T = 2(15)(sin30°)/9.8

T = 15/9.8

T = 1.53 seconds

Hence the time rate of flight is 1.53 seconds

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A beaker of vegetable oil contains a beam of light that is aimed at a surface at an angle of 34 degrees as shown. If the index o

OverLord2011 [107]

Answer:

Angle of reflection of light is 34 degree

Explanation:

As per law of reflection of light we know that

angle of incidence of light = angle of reflection of light

So here we know that

angle of incidence on the surface of oil is given as

$\theta_i = 34 degree$

so we know that

$\theta_i = \theta_r$

so here we can say that reflection angle of light will be same as angle of incidence

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8 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

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3 years ago

A rotating water pump works by taking water in at one side of a rotating wheel, and expelling it from the other side. If a pump

Xelga [282]

Answer:

Explanation:

initial angular velocity, ωo = 0 rad/s

angular acceleration, α = 30.5 rad/s²

time, t = 9 s

radius, r = 0.120 m

let the velocity is v after time 9 s.

Use first equation of motion for rotational motion

ω = ωo + αt

ω = 0 + 30.5 x 9

ω = 274.5 rad/s

v = rω

v = 0.120 x 274.5

v = 32.94 m/s

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