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Vesnalui
4 months ago
7

Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 3.2m long and

has mass 290kg . Burt has mass 33kg and Ernie has mass 32kg, Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it,then Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.
Physics
1 answer:
Softa [3K]4 months ago
8 0

Answer:

x = 0.29 m

Explanation:

It is known that the total external force acting on the mass system equals ZERO,

so the center of mass of the entire system will stay stationary.

We find that

m_1\Delta x_1 + m_2\Delta x_2 + m_3\Delta x_3 = 0

Since Ernie approaches Burt's position, we have:

33 x + 290 x + 32(-3.2 + x) = 0

355 x = 102.4

therefore, we conclude that

x = \frac{102.4}{355}

x = 0.29 m

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On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan
ValentinkaMS [3465]

Answer:

The equivalent distance in kilometers is 4012 ×10^{-6} km.

Explanation:

It's known that 1 millimeter converts to 10^{-3} meters. Then, 1 meter converts to 10^{-3} kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = 10^{-3} m

1 m = 10^{-3} km

Thus, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 ×10^{-6} km.

The distance therefore is 4012 ×10^{-6} km.

5 0
4 months ago
A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0
serg [3582]
Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:
v(t)=0.04t^3-0.06t^2
This polynomial can be factored:
v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
t^2=0;\\ 0.04t-0.06=0
By solving these equations, we identify our zeros:
t_1=0; t_2=\frac{3}{2}
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}
From the table, it's evident that our function is positive when - \infty < t and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
s(t)=0.01t^4 - 0.02t^3
We simply substitute t=12 to calculate the total distance traveled:
s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
a(1)=0.12-0.12=0


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A responder can protect himself/herself from radiation by using shielding as a response action. What materials are best for prot
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