Seconds squared is the time unit of acceleration. It represents the change in distance units per second per second. For example, 3 m/sec² means a distance covering 3 meters in the first second, then 9 meters in the 2nd second, and 37 meters in the third second. (3^1, 3^2, 3^3).
Acceleration is part of Newton's 2nd law: force = mass x acceleration. Units of work: joule = kg·m²/s², and power: watts = kg·m²/s³ all contain accelerations.
Answer:
Explanation:
let force exerted by engine be F.Net force =( F-400)N, applying newton law
F-400 = 1.5 x 10³x18 =27000 ,
F = 27400 N.
velocity after 12 s = 0 + 18 x 12 = 216 m/s
Average velocity = (0 + 216 )/2 = 108 m/s
Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶
b) At 12 s , velocity = 216 m/s
Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.
Answer:
very small solid particles called interstellar dust.
Explanation:
In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.
Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).
Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.
Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω = 
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω = x
Gravity affet everything and it touches nothing.
Hope this helps!
