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3 years ago

What is the primary force that holds things together on an astronomical scale is________?

1 answer:

5 0

The <span>primary force that holds things together on an astronomical scale is "Gravitational Force"

Hope this helps!</span>

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State coulomb's law mathematically

Vsevolod [243]

Coulomb's law is express as:

$\begin{gathered} F=k\frac{q_1q_2}{r^2} \\ \text{ where} \\ k\text{ is Coulomb's constant} \\ q_1\text{ and }q_2\text{ are the charges} \\ r\text{ is the distance between charges} \end{gathered}$

7 0

1 year ago

Can someone help me with 1-2

ira [324]

Can't really plot a graph here for question 1.

2a) The car speeds up from A to B. The car travels at a constant speed from B to C. The car slows down to a stop from C to D.

b) From the graph, at 10 seconds, the car is moving at 20 m/s.

4 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

2 years ago

The machine which turns in a power station

Anna [14]

Answer:

generators

Explanation:

the machine which turns in a power station

4 0

3 years ago

If five joules were required to move a crate in 3.7 seconds, what power was applied?

AleksAgata [21]

Answer:

The answer to your question is 1.35 Watts

Explanation:

Data

Work = W = 5 J

time = t = 3.7 s

Power = P = ?

Formula

Power is a rate in which work is done or energy is transferred over time

P = $\frac{W}{t}$

Substitution

$P = \frac{5}{3.7}$

Result

P = 1.35 W

8 0

3 years ago