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3 years ago

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Which is the truest statement? Standing is health-enhancing because it uses energy. Jogging is not a good way tp enhance your he

alth. Jogging requires less energy than swimming. Short walks dp not use enough energy to be health-enhancing

1 answer:

Setler [38]3 years ago

5 0

A. <span>Standing is health-enhancing because it uses energy.

Because standing can burn a lot of calories.</span>

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Does a mirror work because of refraction

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No. A mirror works because of reflection.

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3 years ago

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How many dots should be drawn in the lewis dot structure for an atom of sodium A)19 B)8 C)4 D)1

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Na is in the first column on the periodic table so therefore it would have 1 valence electron
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2 years ago

You went 100 miles north and 14

skad [1K]

Answer:

82 degrees

Explanation:

consider your staying point to be the center of a circle. this center has the coordinates (0, 0).

the radius of the circle is the distance you walked East (14 miles).

I assume your teacher means as "angle of displacement" the angle between the East-West line going through your starting point and the direct line from your starting point to your current position.

then the 100 miles North is tan(displacement angle)×14.

as it is the same, if you first went North and then East, or the other way around. you end up at the same point, with the same coordinates.

so, again.

100 = 14×tan(angle)

tan(angle) = 100/14 = 50/7 = 7.142857...

the displacement angle is then 82 degrees.

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2 years ago

How do I calculate the equivalent resistance?

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2 years ago

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Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human i

Varvara68 [4.7K]

Answer:

$0.0018833\ \text{m/s}$

Explanation:

$d$ = Distance of Andromeda Galaxy from Earth = $2.54\times 10^7\ \text{ly}$

$t$ = Time taken = $90\ \text{years}$

$c$ = Speed of light = $3\times 10^8\ \text{m/s}$

We have the relation

$t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}$

$c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}$

The required answer is $0.0018833\ \text{m/s}$ .

7 0

2 years ago