1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
Alex places 2 cubes side-by-side on a ramp made of wood. Cube #1 is ice and Cube #2 is wood
zmey [24]

Answer:

Explanation:

The sandpaper block did not move because the forces of friction and gravity were balanced.

6 0
3 years ago
The doppler effect is sensitive only to motion along the line of sight. <br> a. True <br> b. False
4vir4ik [10]
Its vey True trust me on this when I say it is
4 0
3 years ago
The first law of motion applies<br> to what?
lana66690 [7]
First law of motion<span>- sometimes referred to as the </span>law<span> of inertia. An object at rest stays at rest and an object in </span>motion<span> stays in </span>motion<span> with the same speed and in the same direction unless acted upon by an unbalanced force.</span>
3 0
3 years ago
A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat
Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0
3 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0
3 years ago
Other questions:
  • Which scientist first proposed physical laws to mathematically describe the effect of forces on the motions of bodies?
    10·2 answers
  • This problem has been solved! See the answer A proton with a speed of 3.5x10^6 m/s is shot into a region between two plates that
    12·1 answer
  • A 2.40 μC charge is subject to a 3.00 mN force due to an Electric Field. What is the magnitude of the Electric Field at the loca
    15·1 answer
  • A bird flies in the xy-plane with a velocity vector givenby v = (α-βt2)i + γtj with α=2.4m/s,β=1.6m/s3 and γ=4m/s2 . Thepositive
    6·1 answer
  • Whichiscorrect?(A)anopaqueobjectdoesnotallowlighttopassthroughit.(B)an
    7·2 answers
  • Please help..!
    7·1 answer
  • A sound wave has frequency 620 Hz and wavelength 10.5 m. What is the speed of sound waves?
    9·1 answer
  • An electromagnetic wave moves through electric and magnetic fields. How might the movement of the wave be described?
    10·1 answer
  • Am 1 years old UwU (JK 15)
    10·1 answer
  • I do believe that i need help
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!