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Luden [163]
4 years ago
12

Consider an unknown charge that is released from rest at a particular location in an electric field so that it has some initial

electric potential energy. In what direction will the charge move in regards to its potential energy?
A) a positive charge will move to regions of lower potential energy, but a negative charge will move to regions of higher potential energy
B) a charge will move to regions of lower potential energy, regardless of whether it is positive or negative.
C) a negative charge will move to regions of lower potential energy, but a positive charge will move to regions of higher potential energy
D) a charge will move to regions of higher potential energy, regardless of whether it is positive or negative.
E) both positive and negative charges can move toward either higher or lower potential energy
Physics
1 answer:
sineoko [7]4 years ago
4 0

Answer:

b)

Explanation:

If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.

For a positive charge: ΔEp = q*ΔV < 0 (as ΔV < 0)

For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)

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ANEK [815]

Answer:

The coefficient of friction is <u>0.242.</u>

Explanation:

Given

Mass on the table is, m_1 = 4.1 kg

Hanging mass is, m_2 = 2.7 kg

Displacement of the masses is, s = 0.355 m

Initial velocity of the masses is, u=0 m/s

Final velocity of the masses is, v = 1.32 m/s

Acceleration by gravity is g = 9.8 m/s².

The acceleration of the system is calculated using equation of motion and is given as:

a=\frac{v^2-u^2}{2s}\\a=\frac{1.32^2-0}{2\times 0.355}=2.46\ m/s^2

Therefore, a = 2.46 m/s²

Applying Newton's second law on the hanging mass, we get:

m_2\times g-T = m_2\times a\\T = m_2\times (g - a)\\T=2.7\times(9.8 - 2.46)\\T = 19.82\ N

Now, applying Newton's second law on the mass on table, we get:

T - F_f = m_1\times a\\F_f =T - m_1\times a\\F_f = 19.82 -4.1\times 2.46\\F_f = 9.73\ N

The normal force of the table on the mass is given as:

N=m_1\times g=4.1\times 9.8=40.18\ N

The coefficient of friction is the ratio of the frictional force and the normal force and is given as:

\mu=\frac{F_f}{N}=\frac{9.73}{40.18}=0.242

Therefore, the coefficient of friction between the mass on the table and the table is 0.242.

7 0
4 years ago
a 120 ohm resistor, a 60 ohm resistor, and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the c
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Answer:
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Knowing V=ri
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What is not a symptom of sleep apnea?
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Explanation:

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3 years ago
A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an
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52.1-4.184=47.916 You then either add or subtract this from 68.6 the first temperature of the metal. This will give you your total.
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3 years ago
Read 2 more answers
A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im
vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}

\dfrac{1}{v}=\dfrac{7}{690}

v=98.57\ cm

We need to calculate the height of the image

Using formula of height

\dfrac{h'}{h}=\dfrac{-v}{u}

Where, h' = height of image

h = height of object

\dfrac{h'}{4}=\dfrac{-98.57}{-30}

h'=\dfrac{98.57\times4}{30}

h'=13.14\ cm

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}

\dfrac{1}{v}=-\dfrac{10}{299}

v=-29.9\ cm

We need to calculate the height of the image

\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}

h=-\dfrac{29.9\times4}{13.0}

h=-9.2\ cm

The image is virtual and erect.

Hence, This is required solution.

6 0
3 years ago
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