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3 years ago

Define a system who's momentum is observed

Physics

1 answer:

Alenkinab [10]3 years ago

5 0

product of force and perpendicular distance

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If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?

Akimi4 [234]

Answer:

Bow Line

Explanation:

If the wind or current is pushing your boat away from the dock, bow line should be secured first.

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3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.

4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.

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3 years ago

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4 years ago

Matter can be classified as elements, compounds, and mixtures. Choose all of the elements from the following examples of matter.

trapecia [35]

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Explanation:

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3 years ago

A cartridge electrical heater is shaped as a cylinder of length L = 200 mm and outer diameter D = 20 mm. Under normal operating

KengaRu [80]

Answer:

Ts=51.83C

Explanation:

First we calculate the surface area of the cylinder, neglecting the top and bottom covers as indicated by the question

Cilinder Area= A=πDL

L=200mm=0.2m

D=20mm=0.02m

A=π(0.02m)(0.2m)=0.012566m^2

we use the equation for heat transfer by convection

q=ha(Ts-T)

q= heat=2Kw=2000W

A=Area=0.012566m^2

Ts=surface temperature

T=water temperature=20C

Solving for ts

Ts=q/(ha)+T

Ts=2000/(5000*0.012566m^2)+20=51.83C

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3 years ago

In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way i

lara [203]

Answer:

Approximately $1.44\times 10^3 \; \rm N \cdot m^{-1}$ assuming that the spring has zero mass.

Explanation:

Without any external force, a piece of mass connected to an ideal spring (like the chair in this question) will undergo simple harmonic oscillation.

On the other hand, the force constant of a spring (i.e., its stiffness) can be found using Hooke's Law. If the spring exerts a restoring force $\mathbf{F}$ when its displacement is $\mathbf{x}$ , then its force constant would be:

$\displaystyle k = -\frac{\mathbf{F}}{\mathbf{x}}$ .

The goal here is to find the expressions for $F$ and for $x$ . By Hooke's Law, the spring constant would be ratio of these two expressions.

Let $T$ represent the time period of this oscillation. With the chair alone, the period of oscillation is $T = 1.00\; \rm s$ .

For a simple harmonic oscillation, the angular frequency $\omega$ can be found from the period:

$\displaystyle \omega = \frac{2\pi}{T}$ .

Let $A$ stands for the amplitude of this oscillation. In a simple harmonic oscillation, both $\mathbf{F}$ and $\mathbf{x}$ are proportional to $A$ . Keep in mind that the spring constant $k$ is simply the opposite of the ratio between $\mathbf{F}$ and $\mathbf{x}$ . Therefore, the exact value of $A$ shouldn't really affect the value of the spring constant.

In a simple harmonic motion (one that starts with maximum displacement and zero velocity,) the displacement (from equilibrium position) at time $t$ would be:

$\displaystyle \mathbf{x}(t) = A \cos(\omega \cdot t)$ .

The restoring velocity at time $t$ would be:

$\displaystyle \mathbf{v}(t) = \mathbf{x}^\prime(t) = -A\, \omega \sin(\omega\cdot t)$ .

The restoring acceleration at time $t$ would be:

$\displaystyle \mathbf{a}(t) = \mathbf{v}^\prime(t) = -A\, \omega^2 \cos(\omega\cdot t)$ .

Assume that the spring has zero mass. By Newton's Second Law of motion, the restoring force at time $t$ would be:

$\begin{aligned}& \mathbf{F}(t) \\ &= m(\text{chair}) \cdot \mathbf{a}(t) \\&= -m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)\end{aligned}$ .

Apply Hooke's Law to find the spring constant, $k$ :

$\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= -\left(\frac{-m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)}{A\cos(\omega \cdot t)}\right) \\ &= \omega^2 \cdot m(\text{chair}) \end{aligned}$ .

Again, $\omega$ stands for the angular frequency of this oscillation, where

$\displaystyle \omega = \frac{2\pi}{T}$ .

Before proceeding, note how $A$ was eliminated from the ratio (as expected.) Additionally, $t$ is also eliminated from the ratio. In other words, the spring constant is "constant" at all time. That agrees with the assumption that this spring is indeed ideal. Back to $k$ :

$\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= \cdots \\ &= \omega^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{T}\right)^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{1.00\; \rm s}\right)^2 \times 36.4\; \rm kg\end{aligned}$ .

Side note on the unit of $k$ :

$\begin{aligned} & 1\; \rm kg \cdot s^{-2} \\ &= 1\rm \; \left(kg \cdot m \cdot s^{-2}\right) \cdot m^{-1} \\ &= 1\; \rm N \cdot m^{-1}\end{aligned}$ .

6 0

3 years ago

Define a system who's momentum is observed​

Define a system who's momentum is observed