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2 years ago

The electric potential a distance r from a small charge is proportional to what power of the distance from the charge?

Physics

1 answer:

gizmo_the_mogwai [7]2 years ago

3 0

Answer: r^-1

Explanation:

V = (k × Q) / r, where V is electric potential, k is constant and r is distance from charge.

Since Q will not change, it can be considered a constant too.

So, we can write V ∝ 1/r

which is same as V ∝ r^-1

You might be interested in

A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a _____.

notsponge [240]

Answer: The correct answer is "metalloid".

Explanation:

Metal is the material which can conduct electricity as it contains free electrons. It is good conductor of electricity. For example, copper, silver.

Metal can be drawn into wires. This property of material is called ductile.

Metal can be beaten into sheets. This property of material is called malleable.

Non metal is the material which cannot conduct electricity as it does not contain free electrons. They are poor conductor of electricity. For example, oxygen.

Metalloid: It has properties of both metals and non metal. It is electrical conductor. For example, semiconductor- silicon and germanium. But they cannot be pulled into wires.

Therefore, a material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metalloid.

6 0

2 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

8 0

3 years ago

Increasing which two things would increase gravitational potential energy?

sukhopar [10]

Answer:

Mass and height

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about $9.8 m/s2.$

Which is represented as;

$PE_g=mgh$

$PE_g$ stands for gravitational potantial energy,

m stands for mass of object,

g is the gravitational constant and

h is the height.

Here we see that mass of object and height is directly proportional to the gravitational potential energy.

That means increasing in mass and height will result in increasing gravitational potential energy.

7 0

2 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

75 kmh⁻¹

Explanation:

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

2 years ago

What can be defined as the rate at which velocity changes

taurus [48]

Answer:

acceleration

Explanation:

The rate at which velocity changes is the definition of the physical quantity called acceleration, and it is given by the formula: $a=\frac{v_f-v_i}{\Delta t}$

where $\Delta t$ is the time that took to change from the initial velocity $v_i$ to the final velocity $v_f$

4 0

2 years ago

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