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kodGreya [7K]
3 years ago
9

A rotating light is located 13 feet from a wall. The light completes one rotation every 3 seconds. Find the rate at which the li

ght projected onto the wall is moving along the wall when the light's angle is 15 degrees from perpendicular to the wall.
Physics
1 answer:
saveliy_v [14]3 years ago
5 0

Answer:

29.2 ft/s

Explanation:

The distance of the light's projection on the wall

y = 13 tan θ

where θ is the light's angle from perpendicular to the wall.

The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,

Angular speed = w = (2π/3)

w = (θ/t)

θ = wt = (2πt/3)

(dθ/dt) = (2π/3)

y = 13 tan θ

(dy/dt) = 13 sec² θ (dθ/dt)

(dy/dt) = 13 sec² θ (2π/3)

(dy/dt) = (26π/3) sec² θ

when θ = 15°

(dy/dt) = (26π/3) sec² (15°)

(dy/dt) = 29.2 ft/s

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3 years ago
If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
marshall27 [118]

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

5 0
3 years ago
What is the correct order of the layers' density from lowest density to highest?
Orlov [11]

Answer:

C. crust, mantle, core

Explanation:

density increases as you travel from the crust to the inner core

the crust is on top

next is the mantle

and then the core

6 0
3 years ago
Read 2 more answers
A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len
weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}

\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}

\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0
2 years ago
A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal
Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

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Substituting,

v_x = (40)(cos 20^{\circ})=37.6 m/s

And therefore, the range of the projectile is:

d=v_x t = (37.6)(5.0)=188 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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