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3 years ago

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Jack and Jill ran up the hill at 2.8 m/s . The horizontal component of Jill's velocity vector was 2.1 m/s . What was the angle o

f the hill?What was the vertical component of Jill's velocity?

1 answer:

klasskru [66]3 years ago

7 0

Answer: $\theta =41.409 ^{\circ}$

Explanation:

Given

Jack and Jill ran up the hill at 2.8 m/s

Horizontal component of Jill's velocity vector was 2.1 m/s

Let $\theta$ is the angle made by Jill's velocity with it's horizontal component

Therefore

$2.8cos\theta =2.1$

$cos\theta =\frac{2.1}{2.8}$

$cos\theta =0.75$

$\theta =41.409 ^{\circ}$

Vertical velocity is given by

$V_y=2.8sin41.11=1.85 m/s$

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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:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

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

4 0

2 years ago

A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium

Kaylis [27]

Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

w² = $\frac{k}{m}$

to find the constant (k) of the spring, we use Hooke's law with the initial data

F = - kx

where the force is the weight of the body that is hanging

F = W = m g

we substitute

m g = - k x

k = $- \frac{m g}{x}$

we calculate

k = $- \frac{9.8 m}{- 2.6 \ 10^{-2}}$

k = 3.769 10² m

we substitute in the first equation

w² = $\frac{ 3.769 \ 10^2 \ m }{m}$

w = 19.415 rad / s

angular velocity and frequency are related

w = 2πf

f = $\frac{w}{2\pi }$

f = 19.415 / 2pi

f = 3.09 Hz

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