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3 years ago

A particle travels in a straight line with speed v.

Physics

2 answers:

alexandr402 [8]3 years ago

8 0

Answer:

D 60°

Explanation:

Using trigonometry:

- The new speed (v/2) of the particle corresponds to the hypothenuse

- The component of v/4 represents the side adjacent to the angle that we want fo find, $\theta$

So we can write:

$cos \theta = \frac{adjacent}{hypothenuse}=\frac{v/4}{v/2}=\frac{1}{2}$

So we find the angle

$\theta= cos^{-1} (\frac{1}{2})=60^{\circ}$

Elan Coil [88]3 years ago

3 0

Answer:

D 60°

Explanation:

Using trigonometry in the attached image:

$cos\alpha =\frac{v/4}{v/2}$

$cos\alpha =\frac{1}{2}$

$\alpha =cos^{-1} \frac{1}{2}$

Angle=60°

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The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is: