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2 years ago

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If your kinetic energy is different from your work input (either greater or less) throuroughly explain what caused this discrepe

ncy

1 answer:

Romashka-Z-Leto [24]2 years ago

8 0

As per work energy theorem,
Work done by all forces = change in kinetic energy
If kinetic energy is different than your work input, then certainly other forces which are responsible for this discrepancy. These forces can be internal or external.

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A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant

atroni [7]

Answer:

The time constant is $\tau = 1.265 s$

Explanation:

From the question we are told that

the time take to charge is $t = 2.4 \ s$

The mathematically representation for voltage potential of a capacitor at different time is

$V = V_o - e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant

$V_o$ is the potential of the capacitor when it is full

So the capacitor potential will be 100% when it is full thus $V_o =$ 100% = 1

and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

V = 0.85

Hence

$0.85 = 1 - e^{-\frac{2.4}{\tau } }$

$- {\frac{2.4}{\tau } } = ln0.15$

$\tau = 1.265 s$

7 0

3 years ago

The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1

ahrayia [7]

Answer:

The temperature is $T = 168.44 \ K$

Explanation:

From the question ewe are told that

The rate of heat transferred is $P = 13.1 \ W$

The surface area is $A = 1.55 \ m^2$

The emissivity of its surface is $e = 0.287$

Generally, the rate of heat transfer is mathematically represented as

$H = A e \sigma T^{4}$

=> $T = \sqrt[4]{\frac{P}{e* \sigma } }$

where $\sigma$ is the Boltzmann constant with value $\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.$

substituting value

$T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }$

$T = 168.44 \ K$

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3 years ago

Based on Kepler's work, which best describes the orbit If a planet around the Sun?

krek1111 [17]

Answer:

an ellipse with the Sun at one focus or D

Explanation:

Answer for edgenuity

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3 years ago

Which shows the correct lens equation? The inverse of f equals the inverse of d Subscript o Baseline times the inverse of d Subs

musickatia [10]

Answer:

The inverse of f equals the inverse of d Subscript o Baseline plus the inverse of d Subscript I Baseline.

Explanation:

The lens equation shows the relation among focal length of the lens, image distance and object distance. It can be expressed as:

$\frac{1}{f}$ = $\frac{1}{d_{o} }$ + $\frac{1}{d_{I} }$

where: f is the focal length of the lens, $d_{o}$ is the object distance to the lens and $d_{I}$ is the image distance to the lens.

The lens equation can be used to determine the unknown value among the variables f , $d_{o}$ and $d_{o}$ .

6 0

3 years ago

Read 2 more answers

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$

Generally, the equation for Rate of flow of Liquid is mathematically given as

$\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}$

$$

Where dP is pressure difference r is the radius

$\mu$ is the viscosity of water

L is the length of the pipe

$Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}$

$Q=5.2 \mathrm{~mL} / \mathrm{s}$

In $30s the quantity that flows out of the tube

$&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}$

In conclusion, the quantity that flows out of the tube

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

Read more about the flows rate

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1 year ago