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2 years ago

An egg is dropped from a building that is 61 m high.

Physics

1 answer:

Allisa [31]2 years ago

3 0

Answer:

Initial Velocity = 0 m/s

Final Velocity = 34.6 m/s

time = 3.5 s

Explanation:

The initial velocity must be zero since, the egg must be at rest initially, before dropping.

Initial Velocity = 0 m/s

Now, for time we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = Height = 61 m

Vi = Initial Velocity = 0 m/s

g = 9.8 m/s²

t =time = ?

Therefore,

61 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (61 m)(2)/(9.8 m/s²)

t = √(12.45 s²)

t = 3.5 s

Now, for final velocity we will use 1st equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(3.5 s)

Vf = 34.6 m/s

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$c=f\lambda$

Thus the wavelength is given by,

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Substitute 3×10⁸m/s for c and 1120×10³Hz for f.

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3 years ago

A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor

Allisa [31]

Answer:

the impedance of the circuit is 25.7 ohms.

Explanation:

It is given that,

Voltage, V = 50 volts

Frequency, f = 60 Hz

Resistance, R = 25 ohms

Capacitive resistance, $X_C=18\ ohms$

Inductive resistance, $X_L=24\ ohms$

We need to find the impedance of the circuit. It is given by :

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$Z=\sqrt{25^2+(24-18)^2}$

Z = 25.7 ohms

So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.

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Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

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$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

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