The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
a = g = -9.8 
downward (acceleration due to gravity).
By using Suvat equation:
v = u + at
where: v is the velocity at time t
u = 40.0 m/s is the initial velocity
a = g = -9.8 
is the acceleration
To find the time t at which the velocity is v = 20.7 m/s
Therefore,
The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
The complete question is:
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?
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Answer:
At the high temperatures of the inner solar nebula, the small proto-planets were too hot to hold the volatile gases that dominated the solar nebula. These proto-planets were Earth, Mars, Venus, and Mercury.
Explanation:
The materials that accreted into the early Earth were probably added piecemeal, without and particular order. The early earth was very hot from gravitational compression, impacts and radioactive decay; the earth was partially molted. The denser metallic liquids sank to the center of the Earth and less denser silicate liquids rose to the top. In this way the Earth differentiated very quickly into a metallic, mostly iron core and a rocky silicate mantle.
-- loud sounds
-- bright lights
-- strong radio signals
-- Slinkies that can pinch you painfully
-- a tsunami in the ocean
-- earthquakes above Richter 5 or 6
0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
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