Answer:
Explanation:
(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...
4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity
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(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...
d = vt
d = (13.8 m/s)(2 s) = 27.6 m
The water is about 27.6 m below ground.
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* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:
vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2
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If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.
Answer:
option B
Explanation:
It is given that in the tire the recommended pressure is 35 PSI however the current pressure is only 26 PSI which means that pressure in the tire is less than the recommended so the chances of blowout of the tire gets eliminated hence option A is not correct.
Having pressure less in the tire can lead to the Unstable handling of the vehicle.
so correct answer is option B
When the box IS on the shelf 2m above the ground,
its potential energy is
(weight) x (height) = (3 N) x (2 m) = 6 joules .
THAT's the work you have to do, to lift the box up to there.
Answer:
Explanation:
Initially no of atoms of A = N₀(A)
Initially no of atoms of B = N₀(B)
5 X N₀(A) = N₀(B)
N = N₀ 
N is no of atoms after time t , λ is decay constant and t is time .
For A
N(A) = N(A)₀ 
For B
N(B) = N(B)₀ 
N(A) = N(B) , for t = 2 h
N(A)₀ = N(B)₀
N(A)₀ = 5 x N₀(A)
= 5
= 5 
half life = .693 / λ
For A
.77 = .693 / λ₁
λ₁ = .9 h⁻¹
= 5
Putting t = 2 h , λ₁ = .9 h⁻¹
= 5 
= 30.25
2 x λ₂ = 3.41
λ₂ = 1.7047
Half life of B = .693 / 1.7047
= .4065 hours .
= .41 hours .
