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2 years ago

Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?

Physics

1 answer:

Ksenya-84 [330]2 years ago

7 0

Answer:

The answer is " $6\ \frac{m}{s}$ "

Explanation:

The formula for velocity:

$\to \overline{v}={\frac{\Delta x}{\Delta t}}$

$=\frac{6}{1}\\\\=6\ \frac{m}{s}$

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Which of the following is an example of kinetic energy?

tresset_1 [31]

B. Sound, because everything else sits still and sound waves move

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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

How much would you have to eat for your appendix to explode ? I ate so much on thanksgiving and i need to know i’m very scared

dimulka [17.4K]

Answer:

I don't think your appendix can explode because you ate too much honestly. It's not even possible to eat so much that your appendix explodes, and if you're feeling any pain it definitely isn't because your appendix is about to explode, believe me. Also you could just type it into the internet, that'd be a much faster solution.

6 0

3 years ago

Formed through longshore drift a. sea stack b. sandbar c. spit d. headland

ANTONII [103]

<h3>Answer;</h3>

Sand Spit or Spit

<h3>Explanation;</h3>

Long shore drift is the process that occurs when a sheet of water moves on and off the beach, in other words the swash and back swash, thus capturing and transporting sediment on the beach back out to the sea.
Sandbar is normally formed when the sandspit stretches across a bay and connects the two sides. Headland is a high piece of land that extends out onto the sea. Sea stacks on the other hand results from the collapsing of the roof of the arch.

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3 years ago

Read 2 more answers

A learner sets up a circuit to determine the conductivity of certain solids. What piece of apparatus is not required for the nex

Lilit [14]

It is D as u dont need a stop watch aft that

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2 years ago