F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
Answer:
c. detecting the gravitational effect of an orbiting planet (The Wobble"") by looking for the Doppler shifts in the star's spectrum
Explanation:
In a solar system the mass of the star and planets affect each other's orbital movements. The center of gravity of a star and a planet is inside the star. This causes the star to be closer and farther from the Earth at different times. Due to this wobble the star appears to be red shifted when it is farther and blue shifted when it is closer.
When the mass of the planet is high, like a hot Jupiter it causes more wobble i.e., change in radial velocity. This makes it easier to detect the planet. The earliest hot Jupiter found by this method is the planet 51 Pegasi b.
Answer:
Decreases the transparency of the atmosphere to infrared light.
Explanation:
When a large amount of green-house gases are present in the atmosphere, the layer of these gases become opaque to infrared radiation and radiation from the sun get trapped into these gases molecules. These excited molecules radiate this energy into our own atmosphere and that why the temperature of Earth is rising due to the Green-House effect.
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
