How much heat<span> is </span>required<span> to </span>heat 0.1 g<span> of </span>∆hvap<span> =</span>2260 j/g ∆h<span> =</span>340j/g fus iceat−30 ctosteamat 100c?use<span> the </span>approximate values<span>?</span>
Answer:
-6.44 m/s²
Explanation:
Given:
Δx = 60 m
v₀ = 27.8 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (27.8 m/s)² + 2a (60 m)
a = -6.44 m/s²
Answer:
Since incident wave and its reflected part in opposite phase superimpose on each other
So correct answer will be
Option B
Explanation:
Here we know that the wave reflection is done by rigid boundary
So when wave is reflected by the boundary then its phase is reversed by 180 degree
so the reflected wave is in reverse phase from the boundary
so we can superimpose the reflected part with incident wave to dine the resultant wave
So the phenomenon is given as follow
Answer: 6067.5 N
Explanation:
Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.
Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.