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2 years ago

Explain the uses of Electrophorus

Physics

1 answer:

Maslowich2 years ago

8 0

Answer:

Explanation:

Comment

Interesting creature.

These creatures are the only survivors from an era long past, where nothing else has survived. They are eels capable of discharging a large electrical current. A pack of them (they can hunt in groups) can kill just about anything.

Tribes use and tame them to guard the tribe's tame Electrophorus. They are bread and maintained in schools. While the charge is huge, once discharged, these creatures take a very long time to recharge, so they are not entirely invincible.

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Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

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Now, from Charle's Law:

$\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}$

Hence, the correct option is:

A. TA = TB/2.

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2 years ago

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4 years ago

Read 2 more answers

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

We will use the second equation of motion to find out the initial speed of the arrow.

$h= v_it + \frac{1}{2}gt^2\\$

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}$

vi = 9.81 m/s

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

$h= v_it + \frac{1}{2}gt^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

$17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0$

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

t = 1.13 s

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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2 years ago

1. A 4-N force is used to move an object 2 m in 10 s. Which of the following is the power generated while moving the object?

guajiro [1.7K]

Power is the work done per unit time. Therefore,

$\begin{gathered} p=\frac{w}{t} \\ \text{where} \\ w=\text{work done} \\ t=\text{time} \end{gathered}$

Therefore,

$\begin{gathered} \text{work done=fd} \\ f=force=4N \\ d=\text{displacement}=2m \\ t=2\sec s \\ p=\frac{fd}{t}=\frac{4\times2}{10}=\frac{8}{10}=0.8watts \end{gathered}$

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1 year ago

A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?

Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

h = u² sin²θ/2g m

Where,

g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

h = 18² · sin²60 / 2 x 9.8

= 18² x 0.75 / 2 x 9.8

= 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

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3 years ago

Explain the uses of Electrophorus ​

Explain the uses of Electrophorus