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4 years ago

14

Explain why a rapidly spinning gyroscope, mounted in a frame that allows it to rotate in any direction independently of the fram

e’s rotation, tends to maintain its axis pointed in the same direction in space.

1 answer:

REY [17]4 years ago

5 0

conservation of angular momentum. it will tend to keep spinning in same direction ... in linear case, a body would keep going until a force altered its motion

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Help with Number 4 plz <3

vitfil [10]

Nevermind that one i dont know sorry

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4 years ago

A ray diagram is shown. what does the letter x represent?

kobusy [5.1K]

Answer: Letter X in the diagram represent the angle of incidence

Explanation: One of the properties of wave is reflection. But before reflection can occur, there must have been a directed motion of wave particle towards a plane surface. The angle of incidence is the angle between the direction of motion of the wave particle ( for example; light) and the normal to the plane surface.

8 0

4 years ago

Read 2 more answers

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

How many laws are named after sir issac newton

Blizzard [7]

Answer:

Three Laws

Newton's Three Laws of Motion. Sir Isaac Newton: The Universal Law of Gravitation. Sir Isaac Newton and the Unification of Physics & Astronomy.

Explanation:

3 0

3 years ago

For a flourish at the end of her act, a juggler tosses a single ball high in the air. She catches the ball 3.2 s later at the sa

MAXImum [283]

Answer:

15.68 m/s

Explanation:

Given that,

She catches the ball 3.2 s later at the same height from which it was thrown.

When it reaches the maximum height, its height is equal to 0.

It will move under the action of gravity.

$t=\dfrac{2u}{g}$

2 here comes for the time of ascent and descent.

So,

$u=\dfrac{tg}{2}\\\\u=\dfrac{3.2\times 9.8}{2}\\\\u=15.68\ m/s$

So, the initial upward speed of the ball is 15.68 m/s.

6 0

3 years ago