All categories

3 years ago

I NEED HELP PLEASE, THANKS! :) Compare and contrast electric potential energy and electric potential difference?

Physics

1 answer:

nekit [7.7K]3 years ago

8 0

Answer:

Workdone in bringing a unit positive charge from low potential to higher potential against the electric field is electric potential energy.

Electric potential difference is yhe difference of potential energy at the terminals of battery or conductor eire attached to the battery.

Explanation:

You might be interested in

(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from r

Oksi-84 [34.3K]

Answer:

0.321659377 m/s²

1.383138458 m/s

0.321659377 m/s²

0.62667 m/s²

0.7044 m/s² and 27.17°

Explanation:

d = Diameter of rim = 13 in = $13\times 0.0254=0.3302\ m$

r = Radius = $\frac{d}{2}=\frac{0.3302}{2}=0.1651\ m$

$\omega_f$ = Final angular velocity = $80\times\frac{2\pi}{60}=8.37758\ rad/s$

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 4.3 s

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{8.37758-0}{4.3}\\\Rightarrow \alpha=1.94827\ rad/s^2$

Tangential acceleration is given by

$a_t=r\alpha\\\Rightarrow a_t=0.1651\times 1.94827\\\Rightarrow a_t=0.321659377\ m/s^2$

The tangential acceleration of the bug is 0.321659377 m/s²

Tangential velocity is given by

$v=r\omega\\\Rightarrow v=0.1651\times 8.37758\\\Rightarrow v=1.383138458\ m/s$

The tangential velocity of the bug is 1.383138458 m/s

The tangential acceleration is constant which is 0.321659377 m/s²

Centripetal acceleration is given by

$a_c=\frac{a_tt^2}{r}\\\Rightarrow a_c=\frac{0.321659377^2\times 1}{0.1651}\\\Rightarrow a_c=0.62667\ m/s^2$

The centripetal acceleration of the bug is 0.62667 m/s²

The resultant of the acceleration gives us total acceleration

$a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a=\sqrt{0.321659377^2+0.62667^2}\\\Rightarrow a=0.7044\ m/s^2$

Direction is given by

$\theta=cos^{-1}\frac{a_c}{a}\\\Rightarrow \theta=cos^{-1}\frac{0.62667}{0.7044}\\\Rightarrow \theta=27.17^{\circ}$

The magnitude and direction of the acceleration is 0.7044 m/s² and 27.17°

8 0

4 years ago

The Pilot of a plane measures an air velocity of 165Km/h south relative to the plane. An observer on

AVprozaik [17]

The speed obtained by the pilot is not accurate since it is measuring the rate of travel in the wind, true velocity is that compared to the ground. Therefore the speed of the wind is:

v wind = 165 - 145

v wind = 20 km/h<span>

<span>Therefore the wind velocity = 20 km/h against the plane.</span></span>

6 0

4 years ago

The density of un-set concrete is 2.40 x 10 3 kg/m 3 . When a block with volume 0.0708 m 3 is submerged in the liquid, a force o

Alenkinab [10]

Answer:

The answer to your question is weight = 1667 N

Explanation:

Data

density = 2.40 x 10³ kg/m³

volume = 0.0708 m³

Force = 66.7 N

weight of the block = ?

- Formula

density = mass / volume

- Solve for mass

mass = density x volume

-Substitution

mass = (2.40 x 10³)(0.0708)

-Result

mass = 169.92 kg

-Calculate the weight of the body

weight = mass x gravity

-Substitution

weight = 169.92 x 9.82

-Simplification

weight = 1666.9 N ≈ 1667 N

6 0

3 years ago

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined

MrRa [10]

Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

$F_{B}=-5755N$

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

$\frac{mv^{2} }{R}=W+F_{B}\\ F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N$

The minus sign for downward direction

6 0

3 years ago

In a transpiration experiment, the air bubble has an initial volume of 4.33 mL, and an initial pressure of 101.9 kPa. What will

ira [324]

Answer:

Explanation:is in the picture below

8 0

3 years ago