An object is dropped from a height H above the ground. This free-falling object requires 1.02 s to travel the last 39 m before hitting the ground. From what height H above the ground did the object fall?
Answers
Answer:
H = 95.3 m
Explanation:
Let the height from which object fall down is H
so the total time to fall is given as
now it cover 39 m in last 1.02 s of its path
so now we have
by solving above equation
Now the total height is given as
H = 95.3 m
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Answers
Answer: the correct answer is 8*10^-11
inflation theory
steady state theory
big bang theory
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Answer: oscillating universe theory
Explanation:
- The oscillating universe theory is a modification of the famous big bang theory.
- According to this theory, there are successive cycles of expansion and contraction that occurs in the universe and due to the contraction, the universe gets confined into a very small volume which has a large density.
- When the universe gets confined in such a way then, it is said that a 'bounce' will occur and, it would thus oscillate between the big bang and the big crunch.
- Thus, this theory states that for every big bang there is a big crunch.
Final answer:
The Oscillating Universe Theory suggests that for every big bang there is a big crunch, imagining the universe as undergoing endless cycles of expansion and contraction.
Explanation:
The theory that suggests that for every big bang, there is a big crunch is known as the oscillating universe theory. This theory proposes that the universe undergoes endless cycles of expansion (Big Bang) and contraction (Big Crunch). It's like a ongoing cosmic heartbeat, where our universe alternatively expands and contracts.
The other three theories mentioned, the inflation theory, the steady state theory, and the big bang theory, each have their own illustrations of the universe's formation. None of them, however, necessarily involve the concept of a big crunch following a big bang.
Learn more about Oscillating Universe Theory here:
#SPJ12
Answers
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We need to rearrange this so wavelength is the subject. To do this, divide both sides by frequency to get :
Wave speed / frequency = wavelength
327 / 185 = 1.7675 m
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Probably cause 2+2=4 and 4-2=2.
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time to put on the brakes, decelerate at the rate of 100, and come to a
screeching stop after a total distance of exactly 2.7 km.
The speed it reaches while accelerating is exactly the speed it starts decelerating from.
Speed reached while accelerating = (acceleration-1) (Time-1) = .2 time-1
Speed started from to slow down = (acceleration-2) (Time-2) = 1 time-2
The speeds are equal.
.2 time-1 = 1 time-2
time-1 = 5 x time-2
It spends 5 times as long speeding up as it spends slowing down.
The distance it covers speeding up = 1/2 A (5T)-squared
= 0.1 x 25 T-squared = 2.5 T-squared.
The distance it covers slowing down = 1/2 A (T-squared)
= 0.5 T-squared.
Total distance = 2,700 meters.
(2.5 + .5) T-squared = 2,700
T-squared = 2700/3 = 900
T = 30 seconds
The train speeds up for 150 seconds, reaching a speed of 30 meters per sec
and covering 2,250 meters.
It then slows down for 30 seconds, covering 450 meters.
Total time = 180 sec = 3 minutes, minimum.
Observation:
This solution is worth more than 5 points.
Considering train accelerate for 13000s and deaccelerate for 1000s
so total distance covered is = 20cm/s * 13000s + 100cm/s*100s
= (260000 + 10000 )cm
= 270000cm
= 2.7km
So total time taken is - 13100 seconds