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LUCKY_DIMON [66]
3 years ago
5

HELP ASAP PLEASE PHYSICS

Physics
2 answers:
LiRa [457]3 years ago
6 0

Answer:

1. 45.65

2. 1.19x10^19

3. 514OHM

4. 2.45 ( NOT 1.86 OR 1.36)

6. 20.06A

7. 206 ohms

8. 13 Ohms

9. 2 ohms

10. 0.051 ( NOT 0.036 OR  0.089)

Explanation:

I've struggled with both problems 4 and 10

Nataly_w [17]3 years ago
5 0

Answer: did you get the answers?

Explanation:

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iragen [17]
60.3° from due south and 5.89 m/s    For this problem, first calculate a translation that will put John's destination directly on the origin and apply that translation to Mary's destination. Then the vector from the origin to Mary's new destination will be the relative vector of Mary as compared to John. So John is traveling due south at 6.7 m/s. After 1 second, he will be at coordinates (0,-6.7). The translation will be (0,6.7)  Mary is traveling 28° West of due south. So her location after 1 second will be  (-sin(28)*10.9, -cos(28)*10.9) = (-5.117240034, -9.624128762)  After translating that coordinate up by 6.7, you get  (-5.117240034, -2.924128762)  The tangent of the angle will be 2.924128762/5.117240034 = 0.57142693  The arc tangent is atan(0.57142693) = 29.74481039° Subtract that value from 90 since you want the complement of the angle which is now 60.25518961°    So Mary is traveling 60.3° relative to due south as seen from John's point of view.  The magnitude of her relative speed is  sqrt(-5.117240034^2 + -2.924128762^2) = 5.893783 m/s    Rounding the results to 3 significant digits results in 60.3° and 5.89 m/s
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3 years ago
Explain the reason behind why geographers believe the Malthusian theory is not true
g100num [7]

Answer:

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4 years ago
A train travels 94 kilometers in 5 hours, and then 84 kilometers in 3 hours. What is its average speed?
diamong [38]
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3 years ago
A runner starts from rest and in 2 s reaches a speed of 7 m/s. If we assume that the speed changed at a constant rate (constant
PtichkaEL [24]

Answer:

v=3.5\frac{m}{s}

Explanation:

The average speed is defined as:

v=\frac{\Delta x}{\Delta t}

Using the equations for uniformly accelerated motion, we calculate the runner's acceleration:

a=\frac{v_f-v_o}{2}\\a=\frac{7\frac{m}{s}-0\frac{m}{s}}{2s}\\a=3.5\frac{m}{s^2}

Now, we can calculate the distance that the runner travels:

\Delta x=v_0t+\frac{at^2}{2}\\\Delta x=(0\frac{m}{s})(2s)+\frac{3.5\frac{m}{s}(2s)^2}{2}\\\Delta x=7m

Finally, we calculate the runner's average speed:

v=\frac{7m}{2s}\\v=3.5\frac{m}{s}

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