Is the arctic ocean larger than the indian ocean?

A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe

Aleksandr-060686 [28]

T<u>he direction of motion</u> of the person relative to the water is <u>16.7° north of east.</u>

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

$Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

Now, substituting the given information we have:

$alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

$\alpha =Tan(\frac{3.6\frac{m}{s} }{12\frac{m}{s} })^{-1}\\\\\alpha =Tan(0.3)^{-1}=16.69\°(North-East)=16.7\°(North-East)$

Hence, we have that <u>the direction of motion</u> of the person relative to the water is 16.7° north of east.

Have a nice day!

7 0

3 years ago

A 403.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 320.0 K. How much heat must the smelter p

aleksklad [387]

The answer is 2.49 x 10^5 KJ. This was obtained (1) use the formula for specific heat to achieve Q or heat then (2) get the energy to melt the copper lastly (3) Subtract both work and the total energy required to completely melt the copper bar is achieved.

7 0

3 years ago

Read 2 more answers

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

3 years ago

If mars has an orbital period of 1.84 years how far from the sun is it and how

EastWind [94]

<span>141.6 million mi,and idk what u mean by how</span>

6 0

3 years ago

Which pair does NOT have an electric force between them?.

viva [34]

Two neutral objects will not have any electric force of attraction or repulsion between them.

<h3>What is the condition for the electric force between two objects?</h3>

As we know from the electrostatics that whenever there are two charges having a positive charge on one and a negative on the other will attract each other

similarly, if they are having like charges which are both of them having positive or both of them having a negative charge then there will be a force of repulsion between them.

But if both of them or even one of them is neutral then there will not be any electric force between them.

Thus neutral objects will not have any electric force of attraction or repulsion between them.

To know more about the nature of charged particles follow

brainly.com/question/22492496

5 0

2 years ago