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2 years ago

Which of the angles shown in this picture is the angle of incidence? A B

Physics

2 answers:

Dmitry_Shevchenko [17]2 years ago

6 0

The pictures are not attached, therefore, I cannot give a specific choice.
However, I will try to help you out.

The angle of incidence is defined as the angle formed between the ray of light and the normal to the surface that the ray is falling on.

The angle of incidence can be shown in the attached image.

Therefore, for your question, choose the image on which the above description applies.

Hope this helps :)

dsp732 years ago

6 0

B. is the correct answer, hope this helps!

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Imagine a particular exoplanet covered in an ocean of liquid methane. At the surface of the ocean, the acceleration of gravity i

AlladinOne [14]

Answer:

Explanation:

Atmospheric pressure = 7 x 10⁴ Pa

force on a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere = pressure x area

= 7 x 10⁴ x 3.14 x 2 x 2

= 87.92 x 10⁴ N

b )

weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m

Pressure x area

height x density x acceleration of gravity x π r²

= 10 x 415 x 6.2 x 3.14 x 2 x 2

=323168.8 N

c ) Pressure at a depth of 10 m

atmospheric pressure + pressure due to liquid column

= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)

= 7 x 10⁴ + 10 x 415 x 6.2

(7 + 2.57 )x 10⁴ Pa

9.57 x 10⁴ Pa

8 0

3 years ago

An oceanic ridge might be compared to what continental landform? A) delta B) great plains C) mountain range Eliminate D) mountai

Pavlova-9 [17]

A mountain range because an ocean ridge is an underwater mountain hope this helps you

6 0

3 years ago

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

3 years ago

A- 1000 m/s2 Xi-0m Xf-0.75m Vf-?

sleet_krkn [62]

Answer:

The final velocity of the object is, $v_{f}$ = 27 m/s

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, $x_{i}$ = 0 m

The final displacement of the object, $x_{f}$ = 0.75 m

The initial velocity of the object will be, $v_{i}$ = o m/s

The final velocity of the object, $v_{f}$ = ?

The average velocity of the object,

v = ( $x_{f}$ - $x_{i}$ )/ t

= 0.75 / t

The acceleration is given by the relation

a = v / t

1000 m/s² = 0.75 / t²

t² = 7.5 x 10⁻⁴

t = 0.027 s

Using the I equation of motion,

$v_{f}$ = u + at

Substituting the values

$v_{f}$ = 0 + 1000 x 0.027

= 27 m/s

Hence, the final velocity of the object is, $v_{f}$ = 27 m/s

8 0

3 years ago

A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1

attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.

Explanation:

Given:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

To Find:

The average force and momentum

Formulas:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

Solution:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.

5 0

2 years ago