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3 years ago

Write the equation that links current, potential difference, and resistance

1 answer:

4 0

You can write the equation in 3 different ways, depending on which quantity you want to be the dependent variable. Any one of the three forms can be derived from either of the other two with a simple algebra operation. They're all the same relationship, described by "Ohm's Law".

==> Current = (potential difference) / (resistance)

==> Potential difference = (current) x (resistance)

==> Resistance = (potential difference) / (resistance)

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An object is dropped from rest from the top of a 400 m cliff on Earth. If air resistance is negligible, what is the distance the

IrinaVladis [17]

Answer:

176.58 m

Explanation:

t = Time taken = 6 seconds

u = Initial velocity = 0

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 6^2\\\Rightarrow s=176.58\ m$

The object travels 176.58 m from the cliff in 6 seconds.

8 0

3 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

3 years ago

Read 2 more answers

When light in vacuum is incident at the polarizing angle (Brewster's angle) on a certain glass slab, the angle of refraction is

marysya [2.9K]

Answer:

Explanation:

θ( p ) + θ( r ) = 90

θ (r) = angle of refraction and θ ( p ) is polarising angle.

given θ ( r ) = 31.8

θ ( p ) = 90 - 31.8 = 58.2 degree

ii ) Tanθ ( p ) = n ( refractive index ) = Tan 58.2 = 1.6

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3 years ago

A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. ignore frictiona

Blizzard [7]

I just solved similar type of question. You can refer to my solution which I have attached

8 0

3 years ago