As we know by work energy theorem
total work done = change in kinetic energy
so here we can say that wok done on the box will be equal to the change in kinetic energy of the system
initial the box is at rest at position x = x1
so initial kinetic energy will be ZERO
at final position x = x2 final kinetic energy is given as
now work done is given as
so we can say
so above is the work done on the box to slide it from x1 to x2
Answer:
Speed of another player, v₂ = 1.47 m/s
Explanation:
It is given that,
Mass of football player, m₁ = 88 kg
Speed of player, v₁ = 2 m/s
Mass of player of opposing team, m₂ = 120 kg
The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :
V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.
So, the speed of another player is 1.47 m/s. Hence, this is the required solution.
Answer:
At the very beginning it is what states the whole life cycle
Answer: 0.5 seconds
Explanation:
Given that:
Frequency of the George Washington Bridge F = 2.05 Hz
Period T = ?
Recall that frequency is the number of cycles a wave can complete in one second. Hence, frequency is the inverse of period.
i.e F = 1/T
2.05Hz = 1/T
T = 1/2.05Hz
T = 0.488 seconds (Rounded to the nearest tenth as 0.5seconds)
Thus, the period of the George Washington Bridge is 0.5 seconds
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
