The correct answer is <em>"Thermal".</em>
Answer:
0.71121 km/s
Explanation:
= Velocity of planet initially = 54 km/s
= Distance from star = 0.54 AU
= Final velocity of planet
= Final distance from star = 41 AU
As the angular momentum of the system is conserved
When the exoplanet is at its farthest distance from the star the speed is 0.71121 km/s.
You can reason it out like this:
-- The car starts from rest, and goes 8 m/s faster every second.
-- After 30 seconds, it's going (30 x 8) = 240 m/s.
-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s
-- Distance covered in 30 sec at an average speed of 120 m/s
= 3,600 meters .
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The formula that has all of this in it is the formula for
distance covered when accelerating from rest:
Distance = (1/2) · (acceleration) · (time)²
= (1/2) · (8 m/s²) · (30 sec)²
= (4 m/s²) · (900 sec²)
= 3600 meters.
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When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.
When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.
How does he do that ?
By accelerating at 8 m/s². That's about 0.82 G !
He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !
He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.
Answer:
What are the different personal protective equipment?
Personal protective equipment, commonly referred to as “PPE”, is equipment worn to minimize exposure to a variety of hazards. Examples of PPE include such items as gloves, foot and eye protection, protective hearing devices (earplugs, muffs) hard hats, respirators and full body suits. Understand the types of PPE.
Answer:
y = y₀ (1 - ½ g y₀ / v²)
Explanation:
This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i
y = y₀ + v₀ t - ½ g t²
y = y₀ - ½ g t²
for the ball thrown from the ground with initial velocity v₀₂ = v
y₂ = y₀₂ + v₀₂ t - ½ g t²
in this case y₀ = 0
y₂2 = v t - ½ g t²
at the point where the two balls meet, they have the same height
y = y₂
y₀ - ½ g t² = vt - ½ g t²
y₀i = v t
t = y₀ / v
since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height
y = y₀ - ½ g t²
y = y₀ - ½ g (y₀ / v)²
y = y₀ - ½ g y₀² / v²
y = y₀ (1 - ½ g y₀ / v²)
with this expression we can find the meeting point of the two balls
