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yarga [219]
3 years ago
15

For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

rcuit? the current leads the generator EMF by 0.982 radians, what are IT 3 20 the resistance value of the resistor, R?
Physics
1 answer:
MAXImum [283]3 years ago
5 0

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

z = \frac{V_o}{i_o}

now we will have

z = \frac{125}{3.20} = 39.06 ohm

Part b)

Now we also know that

\frac{R}{z} = cos\theta

\theta = 0.982 rad = 56.3 degree

now we have

\frac{R}{39.06} = cos56.3

R = 21.7 ohm

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Answer:

The height of the water slide is 0.878 m

Explanation:

Given that,

Distance = 2.52 m

Suppose Children slide down a friction less water slide that ends at a height of 1.80 m above the pool.

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Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

1.80=0+\dfrac{1}{2}\times9.8\times t^2

t^2=\dfrac{1.80\times2}{9.8}

t=\sqrt{\dfrac{1.80\times2}{9.8}}

t=0.606\ sec

We need to calculate the velocity

Using formula of velocity

v = \dfrac{d}{t}

Put the value into the formula

v=\dfrac{2.52}{0.606}

v=4.15\ m/s

We need to calculate height

Using conservation of energy

\dfrac{1}{2}mv^2=mgh

h=\dfrac{v^2}{2g}

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h=\dfrac{4.15^2}{2\times9.8}

h=0.878\ m

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How Does Earth's gravitational force field act on objects that aren't touching Earth's surface?
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Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

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