Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
Answer:
Number of revolutions=1.532 revolutions
Explanation:
Given data
Distance s=8.0 m
Angular speed a=1.2 rev/s
To find
Number of revolutions
Solution
From the equation of simple motion we not that
So for the number of revolutions she makes is given as
Answer:
-0.912 m/s
Explanation:
When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.
where 
are the mass of the child, the boat and the package, respectively. 
are the velocity of the package and the boat after throwing.
Answer:
138.3 days
Explanation:
Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.
The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.
T^2 is proportional to r^3
That is,
T^2/r^3 = constant
98^2 / 62^3 = T^2 / 78^3
Make T^2 the subject of formula.
T^2 = 98^2 / 62^3 × 78^3
T^2 = 19123.2
T = sqrt ( 19123.2 )
T = 138.2867 days
Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.
Answer:
Torque decreases .
Explanation:
The tape is pulled at constant speed , speed v is constant , so there is
v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .
Let it be α . Let I be moment of inertia of reel .
Reel is in the form of disc
I = 1/2 m r²
I x α = torque
1/2 m r² x α = torque
As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .
