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I am Lyosha [343]
3 years ago
14

You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror

and strikes the wall. The plane of the incident and reflected rays is perpendicular to the wall. The ray from the laser strikes the mirror 30.7 cm from the wall, and the reflected ray strikes the wall 30.1 cm above the mirror. Find the angle of incidence of the laser ray on the mirror.
Physics
1 answer:
joja [24]3 years ago
4 0

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y =  30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle  θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

tan( 90 - θ)  = 30.1 / 30.7

90 - θ = tan^-1 (30.1/30.7)

90 - θ = 44.4345

θ = 45.56 º

the angle of incidence θ is 45.56 º

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A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has
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Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration (a) consist in two components, <em>radial</em> (a_{r}) and <em>tangential</em> (a_{t}), in meters per square second:

a_{r} = \omega^{2}\cdot r (1)

a_{t} = \alpha \cdot r (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

a = \sqrt{a_{r}^{2}+a_{t}^{2}} (3)

Where:

r - Radius of the wheel, in meters.

\omega - Angular speed, in radians per second.

\alpha - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

\omega = \omega_{o}+ \alpha\cdot t (4)

Where:

\omega_{o} - Initial angular speed, in radians per second.

t - Time, in seconds.

If we know that r = 0.1\,m, \alpha = 2\,\frac{rad}{s^{2}}, \omega_{o} = 0\,\frac{rad}{s} and t = 0.60\,s, then the total linear acceleration is:

\omega = \omega_{o}+ \alpha\cdot t

\omega = 1.2\,\frac{rad}{s}

a_{r} = \omega^{2}\cdot r

a_{r} = 0.144\,\frac{m}{s^{2}}

a_{t} = \alpha \cdot r

a_{t} = 0.2\,\frac{m}{s^{2}}

a = \sqrt{a_{r}^{2}+a_{t}^{2}}

a \approx 0.246\,\frac{m}{s^{2}}

The total linear acceleration is approximately 0.246 meters per square second.

3 0
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Are plants carnivores<br>Ture or False​
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Explanation:

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A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
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Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle=40^{\circ}

x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

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Thus range of speed is (10.69, 11.436)

3 0
2 years ago
Water being turned into ice cubes in a freezer is an example of _____.
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7 0
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A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requ
IgorC [24]

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

\frac{\triangle m}{m}=0.004998=0.49985\%

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

Kinetic Energy=\frac{1}{2}mv^2

K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy=\frac{1}{2}mv^2

Now, the relation between energies ratio and masses is:

\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2

\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977

\frac{\triangle m}{m}=0.004998=0.49985\%

3 0
3 years ago
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