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Anna35 [415]
3 years ago
6

A rod of propoer length l0 is at rest in frame S'. it ilies in the x',y' plane and amkes an angle of sin^-1(3/5) What must be th

e value of v if as measured in S the rod is at a 45 degree)
Physics
1 answer:
Blizzard [7]3 years ago
5 0

Answer:

v = 1.98*10^8 m/s

Explanation:

Given:

- Rod at rest in S' frame

- makes an angle Q = sin^-1 (3/5) in reference frame S'

- makes an angle of 45 degree in frame S

Find:

What must be the value of v if as measured in S the rod is at a 45 degree)

Solution:

- In reference frame S'

                x' component = L*cos(Q)

                y' component = L*sin(Q)

- Apply length contraction to convert projected S' frame lengths to S frame:

                x component = L*cos(Q) / γ           (Length contraction)

                y component = L*sin(Q)                  (No motion)

- If the rod is at angle 45° to the x axis, as measured in F, then the x and y components must be  equal:

                L*sin(Q) = L*cos(Q) / γ    

Given:       γ = c / sqrt(c^2 - v^2)

                 c / sqrt(c^2 - v^2) = cot(Q)

                 1 - (v/c)^2 = tan(Q)

                 v = c*sqrt( 1 - tan^2 (Q))

For the case when Q = sin^-1 (3/5)::

                 tan(Q) = 3/4

                 v = c*sqrt( 1 - (3/4)^2)

                 v = c*sqrt(7) / 4 = 1.98*10^8 m/s

 

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yaroslaw [1]

Answer:

Approximately 73\%.

(Assuming that g = \rm 9.81\; m \cdot s^{-2}.)

Explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let m(\text{book}) represent the mass of the book.

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When the book was about to hit the ground, its speed is \rm 40\; m \cdot s^{-1}. Its kinetic energy would be:

\begin{aligned} \text{KE} &= \frac{1}{2} \, m(\text{book}) \cdot v^{2} \cr &= \left(\frac{1}{2} \times 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00\times 10^2\right)\cdot m(\text{book})\end{aligned}.

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The amount of mechanical energy lost in this process would be equal to:

\begin{aligned}&\left(2.943\times 10^3\right) \cdot m(\text{book}) - \left(8.00\times 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143\times 10^3\right)\cdot m(\text{book})\end{aligned}.

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book, m(\text{book}), was eliminated in this process.

\begin{aligned}&\frac{\left(2.143\times 10^3\right)\cdot m(\text{book})}{\left(2.943\times 10^3\right) \cdot m(\text{book})}\times 100\% \cr &= \frac{2.143\times 10^3}{2.943\times 10^3}\times 100\% \cr & \approx 73\%\end{aligned}.

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Answer:

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Explanation:

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The equation can be balance as illustrated below:

H₂S + SO₂ —> H₂O + S

There are 2 atoms of O on the left side and 1 atom on the right side it can be balance by writing 2 before H₂O as shown below:

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