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2 years ago

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Please left home at 8 AM to spend the day at in amusement park. He arrived at the park, which was 150 KM from his house, at 10 A

M. Pace explain that his speed 38 KM/H which was what

1 answer:

antiseptic1488 [7]2 years ago

3 0

Answer:

$v=\frac{150km}{2h}=75\frac{km}{h}$

Explanation:

We can use the equation for the speed

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

where x is the distance and t the time. In this case we know that the time spent was 2 hours and the distance was 150km. By replacing we have

$v=\frac{150km}{2h}=75\frac{km}{h}$

I hope this useful for you

regards

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Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

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a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

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$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

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<h3>a.</h3>

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

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$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

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<h3>b.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

<h3>c.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

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