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3 years ago

12

A charge of 0.14 C is moved from a position where the electric potential is 20 V to a position where the electric potential is 5

0 V. What is the change in potential energy of the charge associated with the change in position?

1 answer:

alukav5142 [94]3 years ago

7 0

The change in electric potential energy is given by:

ΔU = ΔVq

ΔU = change in PE, ΔV = potential difference, q = charge

Given values:

ΔV = 50V - 20V = 30V, q = 0.14C

Plug in and solve for ΔU:

ΔU = 30(0.14)

ΔU = 4.2J

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S ship maneuvers to within 2.50x10^3 m of an islands 1.80x10^3 m high mountain peak and fires a projectile at an enemy ship 6.10

lilavasa [31]

velocity of the projectile is given as

$v = 2.50 \times 10^3 m/s$

now the angle is given as 75 degree

so here we will have

$v_x = 2.50 \times 10^3 cos75 = 0.65 \times 10^3 m/s$

$v_y = 2.50 \times 10^3 sin75 = 2.4 \times 10^3 m/s$

now the time taken to reach the peak is given as

$t = \frac{x}{v_x}$

$t = \frac{2.50 \times 10^3}{0.65 \times 10^3} = 3.86 s$

now the height moved by the projectile in same time is given as

$y = v_y t + \frac{1}{2}at^2$

now we have

$y = (2.4 \times 10^3)(3.86) - \frac{1}{2}(9.81)(3.86)^2$

$y = 9.2 \times 10^3 m$

so distance between the peak and projectile is given as

$d = 9.2 \times 10^3 - 1.80 \times 10^3 = 7.4 \times 10^3 m$

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3 years ago

Which of the following best explains how the Ptolemy’s (geocentric) and Copernicus’s (heliocentric) models of the solar system d

TiliK225 [7]

Answer:

The correct answer is c

Explanation:

In these two different models of movement of the planets

Ptolemy raises the Earth as the center of the solar system

In the Copernicus system, it poses the Sun as the center of the solar system.

Copernicu's system was accepted for giving a simpler and more complete explanation of the problem

The correct answer is c

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4 years ago

Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.

Vadim26 [7]

Heya!!

For calculate aceleration, let's applicate second law of Newton:

$\boxed{F=ma}$

⇒ Being:

→ F = Force = 12 N

→ m = Mass = 3 kg

→ a = aceleration = ?

Lets replace according formula and leave the "a" alone:

$12\ N = 3\ kg * \textbf{a}$

$\textbf{a} = 12\ N / 3\ kg$

$\textbf{a} = 4\ m/s^{2}$

Result:

The aceleration of the object is of <u>4 m/s²</u>

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how many miles can you get on one tank of gas if your tank holds 18 gallons and you get 23 miles per gallon?

andrezito [222]

(18 gallon/tank) x (23 mile/gallon) = <em>414 mile/tank</em>

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An object of mass m is dropped from height h above a planet of mass M and radius R. Find an expression for the objectâ€™s speed

Oksana_A [137]

Answer:

Explanation:

Given

mass of object is m

Mass of planet is M

radius of planet is R

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When it falls on earth with some velocity v

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As Energy is conserved therefore

$=E_2$

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$\frac{1}{2}mv^2=\frac{GMmh}{R(R+h)}$

$v=\sqrt{\frac{2GMh}{R(R+h)}}$

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3 years ago