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aivan3 [116]
3 years ago
15

The brightening of the night sky by light scattered from artificial outdoor lighting is called answer

Physics
1 answer:
kondor19780726 [428]3 years ago
7 0
The brightening of the night sky by light scattered from artificial outdoor lighting is called <span>Light Pollution</span>
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A submarine travels 25 km/h north for 3.2 hours. What is its displacement?
Vika [28.1K]

displ = velocity x time

25 x 3.2 = 75+5 km north.

7 0
3 years ago
Read 2 more answers
100 points!!! Please help!!!
Greeley [361]

Answer:

41°

Explanation:

Kinetic energy at bottom = potential energy at top

½ mv² = mgh

½ v² = gh

h = v²/(2g)

h = (2.4 m/s)² / (2 × 9.8 m/s²)

h = 0.294 m

The pendulum rises to a height of  above the bottom.  To determine the angle, we need to use trigonometry (see attached diagram).

L − h = L cos θ

cos θ = (L − h) / L

cos θ = (1.2 − 0.294) / 1.2

θ = 41.0°

Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.

7 0
3 years ago
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it
fomenos

Answer:68.15m/s

Explanation:

<u><em>Given: </em></u>

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

<u><em>Formula:</em></u>

v₁²=v₁²+2a (x)

<u>Set up:</u>

=\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)

<h2><u><em>Solution:</em></u></h2><h2><u><em>68.15m/s</em></u></h2>

<u />

6 0
3 years ago
5. An acrobat, starting from rest, swings freely on a trapeze of
34kurt

The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;

     a) The maximum speed is v = 4.89 m / s

     b) The maximum height is h = 1.22 m

The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.

               Em = K + U

Let's write the energy in two points.

Starting point. Highest part of the oscillation

            Em₀ = U = m g h

Final point. Lower part of the movement

            Em_f = K = ½ m v²

Energy is conserved.

            Emo = Em_f  

            m g h = ½ m v²

            v² = 2 gh

Let's use trigonometry to find the height, see attached.

         h = L - L cos θ

         h = L (1- cos θ)

They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.

         h = 3.7 (1- cos 48)

          h = 1.22 m

this  is the maximum height of the movement.

Let's calculate the velocity.  

          v= \sqrt{2 \ 9.8 \ 1.22}  

          v = 4.89 m / s

In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;

     a) The maximum speed is v = 4.89 m / s

     b) The maximum height is h = 1.22 m

Learn more here: brainly.com/question/13010190

5 0
3 years ago
To view an enlarged upright image of an object through a simple magnifier, where must the object be located?.
Andre45 [30]

Answer:within the focal length of the lens, provided the focal length is shorter than the near point distance.

Explanation:Hope it helps

6 0
2 years ago
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