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3 years ago

All satellite eventually lose their orbital energy and plummet to earth because of______.

Physics

1 answer:

Softa [21]3 years ago

8 0

Answer:

gravity

Explanation:

Even when satellites are thousands of miles away, Earth's gravity still tugs on them. Gravity—combined with the satellite's momentum from its launch into space—cause the satellite to go into orbit above Earth, instead of falling back down to the ground.

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If a skydiver jumps out of a plane horizontally (in other words with no initial vertical velocity), then what will her vertical

erastovalidia [21]

The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.

<h3>Time of motion of the girl</h3>

The time of motion of the girl is calculated as follows;

h = vt + ¹/₂gt²

where;

v is initial vertical velocity = 0
t is time of motion
g is acceleration due to gravity

Substitute the given parameters and solve for time of motion;

50.8 = 0 + ¹/₂(9.8)t²

2(50.8) = 9.8t²

101.6 = 9.8t²

t² = 101.6/9.8

t² = 10.367

t = √10.367

t = 3.22 seconds

<h3>Final vertical velocity of the skydiver</h3>

vf = vi + gt

where;

vi is the initial vertical velocity = 0

vf = 0 + 9.8(3.22)

vf = 31.56 m/s

Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.

Learn more about vertical velocity here: brainly.com/question/24949996

#SPJ1

4 0

2 years ago

Charge X has twice as much charge as particle Y. The two charges are placed near each other. Compared to the force on particle X

Art [367]

The force on charge Y is the same as the force on charge X

Explanation:

We can answer this problem by applying Newton's third law of motion, which states that:

"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by

$F=k\frac{q_x q_y}{r^2}$ (1)

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_x, q_y$ are the two charges

r is the separation between the two charges

According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

Any five chapter 9 institutions found in the constitution of south africa

navik [9.2K]

Five institutions from the chapter 9 found in the constitution of south Africa are :
- The Public Protector
- The south African Human right commission
- The commission of Gender Equality
- The Auditor- General
- The Independent Electoral Commission

Hope this helps

7 0

3 years ago

If the crate shown here is moving at a constant speed in a straight line and the force applied is 310 N, what is the magnitude o

larisa86 [58]

Answer:

$f_k = 310N$

the answer is A.

Explanation:

Using the laws of newton:

∑F = ma

where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.

Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:

∑F = m(0)

∑F = 0

It means that:

F - $f_k$ = 0

where F is the force applied and $f_k$ is the friction force. Replacing the value of F, we get:

310N - $f_k$ = 0

Finally, solving for $f_k$ :

$f_k = 310N$

8 0

3 years ago