Explanation:
Orbital speed= 2pi x radius / time period
=2pi x 1.5x10^11 / 365.25
=2.58x10^9m/day
687 days
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The spring will come to rest 4.9 m below the natural length
Explanation:
The mass-spring system will come to rest when the restoring force on the spring (pulling upward) balances the weight of the mass (pulling downward). Mathematically, this can be written as
where
k is the spring constant
x is the elongation of the spring
m is the mass
g is the acceleration of gravity
In this problem, we have:
is the mass
is the acceleration of gravity
is the spring constant
Solving the equation for x,
Therefore, the spring will come to rest 4.9 m below the natural length.
Learn more about forces:
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Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Answer:
-10
Explanation:
The rectangular area is the base bbb times height hhh.
\begin{aligned}\Delta x&= A \\\\ &=(2\,\text s)\left (-5\,\dfrac{\text m}{\text s}\right ) \\\\ &= -10\,\text m\end{aligned}
Δx
=A
=(2s)(−5
s
m
)
=−10m
The possum displaces 10\,\text m10m10, start text, m, end text to the left.
