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alukav5142 [94]

2 years ago

9

A 42.0 kg projectile is launched horizontally from the top of a cliff, 50 meters above the ground. The projectile’s initial spee

d is 30 m/s. What is the velocity right before it hits the ground

1 answer:

yanalaym [24]2 years ago

5 0

Answer:

Velocity of the projectile just before it will hit the surface is 43.4 m/s

Explanation:

In x direction there is no acceleration

so horizontal velocity of the projectile is always constant

So we have

$v_x = 30 m/s$

now in Y direction its motion is free fall motion under gravity

so we have

$v_y^2 = 0^2 + 2(9.81)(50)$

$v_y = 31.3 m/s$

so we have

$v = \sqrt{v_y^2 + v_x^2}$

$v = \sqrt{30^2 + 31.3^2}$

$v = 43.4 m/s$

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Answer:

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Answer:

A. $U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

$C=\dfrac{\epsilon A}{d}$

$C$ is the capacitance, $A$ is the common plate area, $d$ is the plate separation and $\epsilon$ is the permittivity of the material between the plates.

For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,

$U = \dfrac{CV^2}{2}$

A. Substituting for $C$ in $U$ ,

$U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. When the distance is $3d$ ,

$U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}$

$U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. When the distance is restored but with a dielectric material of dielectric constant, $K$ , inserted, we have

$U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

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2 years ago

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?

Elan Coil [88]

Answer:

Time period, $T=3.05\times 10^{-5}\ s$

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

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$T=\dfrac{1}{f}$

T is the time period of the crystal's motion.

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$T=\dfrac{1}{32768}$

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