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A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = $4.5 \times 10^{-5} C$

Electric Field strength (E) = $2.0 \times 10^{4} N/C$

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

$U = q \times E \times d$

So now insert the values to find the answer

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

On further solving

$U = 0.027 J$

8 0

2 years ago

An object is thrown directly upwards from the ground at a velocity of 9ms. Recalling that the acceleration due to gravity is −gm

olga55 [171]

Answer:

Explanation:

Given

Object is thrown with a velocity of $u=9\ m/s$

Acceleration due to gravity is -g (i.e. acting downward)

Vertical distance traveled by object is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

at maximum height final velocity is zero

$0-(9)^2=2\times (-g)\times (s)$

$s=\frac{81}{2g}=\frac{40.5}{g}\ m$

time taken to reach maximum height

using

v=u+at

0=9-gt

$t=\frac{9}{g}\ s$

6 0

3 years ago

Light waves travel in straight lines in all directions from the source of the light, unless

Ainat [17]

A. the medium through which the light travels changes.

Explanation:

Light waves will continue to travel in a straight line in all directions from their source unless the medium through which the light travels changes.

A change in medium causes light to exhibit different properties. Also, when light hits an obstacle, they can be diffracted.

The way light travels on crossing a boundary differs.
At the boundary between two medium, light can either be reflected back or refracted when they cross the medium
This will cause the light rays to bend towards or away from the normal depending on the properties of the medium.

Learn more:

Refraction brainly.com/question/12370040

#learnwithBrainly

6 0

3 years ago

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

7 0

3 years ago

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actu

Juliette [100K]

$\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}$

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

${\bold{\blue{GIVEN}}}$

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

${ \bold{\green{To \: Find}}}$

REAL DEPTH OF THE OBJECT.

${\red{FORMULA \: \: USED }}$

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth }$

$\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}$

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } \\ \\ 1.3 = \frac{Real \: Depth}{7.7 \: cm} \\ \\ 1.3 \times 7.7 = Real \: Depth \\ \\ 10.01 \: \: cm = Real \: Depth$

3 0

2 years ago