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3 years ago

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For an independent study project, you design an experiment to measure the speed of light. You propose to bounce laser light off

a mirror that is 51.0 km due east and have it detected by a light sensor that is 135 m due south of the laser. The first problem is to orient the mirror so that the laser light reflects off the mirror and into the light sensor. Determine the angle that the normal to the mirror should make with respect to due west.

1 answer:

Gekata [30.6K]3 years ago

4 0

Answer:

4.55 minute

Explanation:

51 km = 51000 m

the ray of light travels due east 51000 m , bounces back from reflecting mirrors . The reflected light gets focused on a point 135 m south . Let the angle between incident light and reflected light be θ.

tanθ = 135 / 51000

θ = .1516 degree

= .1516 x 60 minute

= 9.09 minute

angle between incident ray ( east - west ) and normal = 9.09 / 2

= 4.55 minute Ans

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A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 41.0-gram mass is attached at the 23.0-cm

aev [14]

Answer:

mass of the meter stick=0.063 kg

or

mass of the meter stick=63.3 g

Explanation:

Given data

m₁=41.0g=0.041kg

r₁=(39.2 - 23)cm

r₂=(49.7 - 39.2)cm

g=9.8 m/s²

To find

m₂(mass of the meter stick)

Solution

The clockwise and counter-clockwise torques must be equal if the meter stick is in rotational equilibrium

$Torque_{cw}=Torque_{cw}\\F_{1}r_{1}=F_{2}r_{2}\\ m_{1}gr_{1}=m_{2}gr_{2}\\(0.041kg)(9.8m/s^{2} )(0.392m-0.23m)=m_{2}(9.8m/s^{2})(0.497m-0.392m)\\0.0651N.m=1.029m_{2}\\m_{2}=0.063 kg\\or\\m_{2}=63.3g$

5 0

3 years ago

Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet of

Harrizon [31]

Answer:

<em>a) 6.57 m/s</em>

<em>b) 53.75 J </em>

<em>c) 6.37 m/s</em>

<em>d) -98.297 J</em>

Explanation:

mass of player = $m_{p}$ = 117.5 kg

speed of player = $v_{p}$ = 6.5 m/s

mass of ball = $m_{b}$ = 0.43 kg

velocity of ball = $v_{b}$ = 26.5 m/s

Recall that momentum of a body = mass x velocity = mv

initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s

initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s

initial kinetic energy of the player = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.5^{2}$ = 2482.187 J

a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.

for this first case that they travel in the same direction, their momenta carry the same sign

$m_{p}$ $v_{p}$ + $m_{b}$ $v_{b}$ = ( $m_{p}$ + $m_{b}$ )v

where v is the final velocity of the player.

inserting calculated momenta of ball and player from above, we have

763.75 + 11.395 = (117.5 + 0.43)v

775.145 = 117.93v

v = 775.145/117.93 = <em>6.57 m/s</em>

b) the player's new kinetic energy = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.57^{2}$ = 2535.94 J

change in kinetic energy = 2535.94 - 2482.187 = <em>53.75 J gained</em>

c) if they travel in opposite direction, equation becomes

$m_{p}$ $v_{p}$ - $m_{b}$ $v_{b}$ = ( $m_{p}$ + $m_{b}$ )v

763.75 - 11.395 = (117.5 + 0.43)v

752.355 = 117.93v

v = 752.355/117.93 =<em> 6.37 m/s</em>

d) the player's new kinetic energy = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.37^{2}$ = 2383.89 J

change in kinetic energy = 2383.89 - 2482.187 = -98.297 J

that is <em>98.297 J lost</em>

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4 years ago

A car traveling at 25 m/s starts to decelerate steadily. It comes to a complete stop in 12 seconds. What is its acceleration?

never [62]

Assume it is in uniform deceleration.
u=25
v=0
s=x
a=?
t=12

v=u+at
0=25+a(12)
a=-2.08(3sig fig)

decelerarion of 2.08 ms^(-2)/
acceleration=-2.08ms^(-2)

6 0

3 years ago

Read 2 more answers

Water is flowing through a 45° reducing pipe bend at a rate of 200 gpm and exits into the atmosphere (P2 = 0 psig). The inlet to

Nataliya [291]

Answer:

F1=177.88 Newtons

Explanation:

Let's start with the Bernoulli's equation:

$P_{1} + \frac{1}{2}\beta V_{1} ^{2} + \beta gh_{1} =P_{2} + \frac{1}{2}\beta V_{2} ^{2} + \beta gh_{2}$

Where:

P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.

From the Bernoulli's equation and assuming that h is constant and P2 is zero (from the data), we have:

$P_{1} + \frac{1}{2}\beta V_{1} ^{2} = \frac{1}{2}\beta V_{2} ^{2}$

As we know, P1 must be equal to $\frac{F_{1} }{A_{1}}$ , so, replacing P1 in the equation, we have:

$P_{1} = \frac{F_{1}}{A_{1}} = \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2})$

And

$F_{1} = {A_{1}} ( \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2}))$

Now, let's find the velocity to replace the values on the expression:

We can express the flow in function of velocity and area as $Q = V A$ , where Q is flow, V is velocity and A is area. As the same, we can write this: $Q_{1} = V_{1} A_{1}\\Q_{2} = V_{2} A_{2}$ . In the last two equations, let's clear Velocities.

$V_{1} = \frac{Q_{1}}{A_{1}}\\V_{2} = \frac{Q_{2}}{A_{2}}$

and replacing V1 and V2 on the last equation resulting from Bernoulli's (the one that has the force on it):

$F_{1} = {A_{1}} ( \frac{1}{2}\beta((\frac{Q_{2}}{A_{2}})^{2} - (\frac{Q_{1}}{A_{1}})^{2}))$

First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.

$D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg$

Replacing the respective values in this last expression, we obtain:

F1 = 177.88 N

3 0

3 years ago

Two volumes of William Shakespeare stand on a bookshelf next to each other: volume one, then volume two. Each volume is 4 cm thi

Setler79 [48]

Answer:

L = 7 [cm]

Explanation:

To solve this problem we must analyze each of the distances mentioned and take into account the number of covers and thicknesses of these.

The worm crosses through the sheets of the first book, this distance can be determined by the following length analysis.

4 = P + 2*C

Where:

P = thicknesses of the pages [cm]

C = thicknesses of each cover [cm]

P = 4 - 2*(0.5)

P = 3 [cm]

The distance crossed was:

L = P + 2C + P "the pages of the first book + 2 covers + the pages of the second book"

L = 3 + (2*0.5) + 3

L = 7 [cm]

7 0

3 years ago