a light beam that hits a mirror at an angle of 36°... what is the Angle of Incidence? Angle of Reflection?

Prove that s=ut+½at²

katrin2010 [14]

Explanation:

Distance travelled = Area under the line

= ut + ½ (v-u)t

Acceleration (a) = (v-u)/t and so (v-u) = at

Therefore,

Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²

Thus,proved.

3 0

2 years ago

A student hears a police siren. What would change the frequency that the student hears? Check all that apply.

ser-zykov [4K]

<span>A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source.

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.</span>

4 0

3 years ago

Read 2 more answers

A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 11.0

kvv77 [185]

The velocity of the trainee is 29 m/s or 0.42 rev/s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Centripetal Acceleration of circular motion could be calculated using following formula:

$\large {\boxed {a_s = v^2 / R} }$

<em>a = centripetal acceleration ( m/s² )</em>

<em>v = velocity ( m/s )</em>

<em>R = radius of circle ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

<u>Unknown:</u>

Velocity of Rotation = v = ?

<u>Solution:</u>

$F = ma$

$F = m\frac{v^2}{R}$

$7.80w = m\frac{v^2}{R}$

$7.80mg = m\frac{v^2}{R}$

$7.80g = \frac{v^2}{R}$

$7.80 \times 9.8 = \frac{v^2}{11.0}$

$v^2 = 840.84$

$v \approx 29 ~m/s$

$\omega = \frac{v}{R}$ → in rad/s

$\omega = \frac{v}{2 \pi R}$ → in rev/s

$\omega = \frac{29}{2 \pi \times 11.0}$

$\omega \approx 0.42 ~ rev/s$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Uniform Circular Motion : brainly.com/question/2562955
Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

6 0

3 years ago

Read 2 more answers

A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from

MAVERICK [17]

Answer:

The acceleration is 3.16x10¹⁷ m/s².

Explanation:

First, we need to find the magnitude of the Coulombs force (F):

$|F| = \frac{Kq_{1}q_{2}}{d^{2}}$

<u>Where</u>:

K is the Coulomb constant = 9x10⁹ Nm²/C²

q₁ is the charge = 20x10⁻⁹ C

q₂ is the electron's charge = -1.6x10⁻¹⁹ C

d is the distance = 1.0 cm = 1.0x10⁻² m

$|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N$

Now, we can find the acceleration:

$a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2}$

Therefore, the acceleration is 3.16x10¹⁷ m/s².

I hope it helps you!

7 0

3 years ago

Describe what a homogeneous electric field is, and where we find such

slega [8]

Answer:

no one really answers on here anymore but can u click on my profile and help me with my question

Explanation:

the answer is A homogeneous electric field has the same magnitude and direction at any place. A good example of a homogeneous field is the field between two charged metal plates. The field strength depends on the voltage U and the plate distance d.

7 0

2 years ago

a light beam that hits a mirror at an angle of 36°... what is the Angle of Incidence? Angle of Reflection?​

a light beam that hits a mirror at an angle of 36°... what is the Angle of Incidence? Angle of Reflection?