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3 years ago

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Maria listed four important points about Einstein’s special theory of relativity. 1) applies to any object moving more slowly th

an the speed of light
2) applies to objects moving in a straight path
3) describes the relationship between time and space
4) applies to uniformly moving objects? Which best describes how Maria can correct her error?

ANSWER A (change the first point to “applies to objects moving almost at the same speed of light”)

2 answers:

DiKsa [7]3 years ago

6 0

<span> applies to any object moving more slowly than the speed of light</span>

kobusy [5.1K]3 years ago

4 0

Answer:

(Change the first point to “applies to objects moving almost at the speed of light.” ) or A

Explanation:

This is the Answer on Edge

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3 years ago

The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from

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Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) $V_{max}$ the car has is:

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And the car's average acceleration $a_{ave}$ is:

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Since:

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Where:

$V_{f}=V_{max}=90.74 m/s$ is the car's final speed (top speed)

$V_{o}=0 m/s$ because it starts from rest

$\Delta t$ is the time it takes to reach the top speed

Finding this time:

$\Delta t=\frac{V_{f}-V_{o}}{a_{ave}}$ (2)

$\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}}$ (3)

$\Delta t=t_{1}=101.95 s$ (4)

Now we have to find the distance $d$ the car traveled at this maximum speed with the following equation:

$V_{f}^{2}=V_{o}^{2} + 2a_{ave} d$ (5)

Isolating $d$ :

$d=\frac{V_{f}^{2}}{2a_{ave}}$ (6)

$d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})}$ (7)

$d=4625.70 m$ (8)

On the other hand, we know the total distance $D$ traveled by the car is:

$D=12.42 miles = 19988.052 m$

Hence the remaining distance is:

$d_{remain}=D-d=19988.052 m - 4625.70 m$ (9)

$d_{remain}=15362.35 m$ (10)

So, we can calculate the time $t_{2}$ it took to this car to travel this remaining distance $d_{remain}$ at its top speed $V_{max}$ , with the following equation:

$V_{max}=\frac{d_{remain}}{t_{2}}$ (11)

Isolating $t_{2}$ :

$t_{2}=\frac{d_{remain}}{V_{max}}$ (12)

$t_{2}=\frac{15362.35 m}{90.74 m/s}$ (13)

$t_{2}=169.45 s$ (14)

With this time $t_{2}$ and the value of $t_{1}$ calculated in (4) we can finally calculate the total time $t_{TOTAL}$ :

$t_{TOTAL}=t_{1}+ t_{2}$ (15)

$t_{TOTAL}=101.95 s + 169.45 s$ (16)

$t_{TOTAL}=271.4 s s$

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Answer:

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