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xeze [42]
3 years ago
8

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

the electric potential energy of the point charge?
× 10–4 J

What is the electric potential energy of the point charge at 1.3 m?


× 10–3 J
Physics
2 answers:
Afina-wow [57]3 years ago
4 0

The electric potential energy of the point charge is 4.

the electric potential energy of the point charge at 1.3 m is 5.2.

Juliette [100K]3 years ago
4 0

Answer:

Part 1)

U = 4 \times 10^{-4} J

Part 2)

U = 5.2 \times 10^{-3} J

Explanation:

Electric field due to rubber rod is given by

E = 1.0 \times 10^3 N/c

now we also know that

V = E. r

V = (1.0 \times 10^3)(0.10)

V = 1.0 \times 10^2

now potential energy is given as

U = qV

U = (4 \mu C)(1.0 \times 10^2)

U = 4 \times 10^{-4} J

Similarly the electric potential at distance r = 1.3 m

so we have

V = (1.0 \times 10^3)(1.3)

V = 1.3 \times 10^3 Volts

now potential energy is given as

U = (4\mu C)(1.3 \times 10^3)

U = 5.2 \times 10^{-3} J

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Which of the following objects cannot make a shadow? Broken piece of glass from a window/wooden pane of the same window? Explain
Lorico [155]

Answer:

broken pieces of glass from a window can't form a shadow.

Explanation:

the reason is that shadow is formed only when light rays hits an opaque object, which doesn't let light to pass through it, but glass is a transparent object, hence light rays passes through it forming no shadow..

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4 0
3 years ago
Instantaneous speed is measured
VMariaS [17]

Answer:

C. At a particular instant

Explanation:

Speed is the defined as the ratio between the distance covered by an object and the time taken:

v=\frac{d}{t}

where d is the distance and t the time.

However, there are two possible measurements of speed:

- Average speed: this is the speed measured over a non-zero time interval (for example: a car moving 100 metres in 5 seconds; its average speed is

v=\frac{100 m}{5 s}=20 m/s

- Instantaneous speed: this is the speed of an object measured at a particular instant in time, so for a time interval that tends to zero. So, in the previous example, the average speed is 20 m/s but the instantaneous speed of the car at various instants of time can be different from that value.

7 0
3 years ago
Please help!!!!
Dafna11 [192]

The intensity of the electric field is 30,000 N/C

Explanation:

The strength of the electric field produced by a single-point charge is given by the equation

E=k\frac{q}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

q=3\cdot 10^{-9}C is the magnitude of the charge

r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity

Substituting, we find:

E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system mo
viva [34]
Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

m_{1} * v_{1} +  m_{2} * v_{2} =  m_{1} * v'_{1} +  m_{2} * v'_{2}

In case of perfectly inelastic collision v'1 and v'2 are same.

We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁

Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg

Mass of second mud ball is 6kg.
7 0
3 years ago
A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres
Lemur [1.5K]

Answer:

V_2 = 0.125 m^3

Work done =  = 5 kJ

Explanation:

Given data:

volume of nitrogen v_1 = 0.08 m^3

P_1 = 150 kPa

T_1 = 200 degree celcius = 473 Kelvin

P_2 = 80 kPa

Polytropic exponent n = 1.4

\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}

putting all value

\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}

\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS

polytropic process is given as

P_1 V_1^n = P_2 V_2^n

150\times 0.08^{1.4} = 80 \times V_2^{1.4}

V_2 = 0.125 m^3

work done = \frac{P_1 V_1 -P_2 V_2}{n-1}

= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}

                  = 5 kJ

4 0
3 years ago
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