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nignag [31]
3 years ago
8

Please help!!!!!!!!!!

Physics
1 answer:
OleMash [197]3 years ago
4 0

Answer:

B (student 1)

Explanation:

...

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Weekend A<br> Assignment<br> Differentiate between forced and damped oscillation
4vir4ik [10]

Answer:

A damped oscillation means an oscillation that fades away with time while Forced oscillations occur when an oscillating system is driven by a periodic force that is external to the oscillating system.

Explanation:

Damping is the reduction in amplitude (energy loss from the system) due to overcomings of external forces like friction or air resistance and other resistive forces. ... When a body oscillates by being influenced by an external periodic force, it is called forced oscillation.

<h2><em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>helped</u></em><em><u> </u></em></h2>

<em><u>Welcome</u></em><em><u> </u></em>

6 0
3 years ago
A commercial jet liner takes off with an average acceleration of 3 g. How long does it take to reach the end of its runway which
ANEK [815]

Answer:

The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

Explanation:

Given;

average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²

distance traveled by the commercial Jet liner, s = 1542 m

The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;

s = ut + ¹/₂at²

where;

u is the initial velocity of the commercial Jet liner = 0

s = 0 + ¹/₂at²

s = ¹/₂at²

2s = at²

t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2 \times 1524}{29.4} } \\\\t = 10.18 \ s

Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

6 0
3 years ago
Diffinition of nebula
GrogVix [38]

nebulae, nebulæ, or nebulas) is an interstellar cloud of dust, hydrogen, helium and other ionized gases. Originally, nebula was a name for any diffuse astronomical object, including galaxies beyond the Milky Way. The Andromeda Galaxy, for instance, was once referred to as the Andromeda Nebula (and spiral galaxies in general as "spiral nebulae") before the true nature of galaxies was confirmed in the early 20th century by Vesto Slipher, Edwin Hubble and others.

Most nebulae are of vast size, even hundreds of light years in diameter.[3] Although denser than the space surrounding them, most nebulae are far less dense than any vacuum created on Earth – a nebular cloud the size of the Earth would have a total mass of only a few kilograms. Many nebulae are visible due to their fluorescence caused by the embedded hot stars, while others are so diffuse they can only be detected with long exposures and special filters. Some nebulae, are variably illuminated by T Tauri variable stars. Nebulae are often star-forming regions, such as in the "Pillars of Creation" in the Eagle Nebula. In these regions the formations of gas, dust, and other materials "clump" together to form denser regions, which attract further matter, and eventually will become dense enough to form stars. The remaining material is then believed to form planets and other planetary system objects.

The range of objects called nebula are very diverse, have diverse origins, and final ends.

Contents <span> [hide] </span><span><span>1Observational history</span><span>2Formation</span><span><span>3Types of nebulae</span><span><span>3.1Classical types</span><span>3.2Diffuse nebulae</span><span><span>3.3Planetary nebulae</span><span>3.3.1Protoplanetary nebula</span></span><span>3.4Supernova remnants</span></span></span><span><span>4Notable named nebulae</span><span>4.1Nebula catalogs</span></span><span>5See also</span><span>6References</span><span>7<span>External links

</span></span></span>

7 0
4 years ago
a child rides her bicycle in her level drive way. what form of energy does she possess A. Elastic B. Nuclear C. Potential D. Kin
Bad White [126]

Answer:

Kinetic Energy

Explanation:

Kinetic energy is energy due to motion.

6 0
3 years ago
Read 2 more answers
Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse
777dan777 [17]

Answer:

      Δy= 5,075 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon is

      d sin θ = (m + ½) λ

As the observation is on a distant screen

     tan θ = y / x

     tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

          tan θ = sin θ

Let's replace

     d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance  

     y₁ = (m + ½) λ x / d

     m = 1

      y₁ = 3/2 λ x / d

Let's use m = 1, we look for the following interference,

             m = 2

             y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

           Δy = λ x / d (5/2 -3/2)

           Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

           Δy= 5,075 10⁻⁶ m

8 0
3 years ago
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