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4 years ago

1. Speed is a measure of

Physics

1 answer:

antiseptic1488 [7]4 years ago

5 0

Answer:

distance traveled over time passed

Explanation:

Speed is a measure of the distance that an object travels over the time passed to travel such distance.

The speed of an object is defined as follows:

$speed=\frac{distance}{time} \\s=\frac{d}{t}$

where $d$ is the distance and $t$ is the time

thus speed is the division of the total distance that is traveled in a certain time.

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A simple harmonic oscillator of amplitude A has a total energy E.

jeka57 [31]

Answer:

a) $K=E-\frac{kA^2}{18}$

b) $U=\frac{kA^2}{18}$

Explanation:

The law of conservation of mechanical energy states that total mechanical energy remains constant during oscillation. Mechanical energy is defined as the sum of kinetic energy and potential energy:

$E=U+K\\E=\frac{kx^2}{2}+\frac{mv^2}{2}$

a) The position is one-third the amplitude. So, we have $x=\frac{1}{3}A$ . Replacing and solving for K

$E=\frac{k(\frac{1}{3}A)^2}{2}+K\\E=\frac{kA^2}{18}+K\\K=E-\frac{kA^2}{18}$

b) The potential energy is defined as:

$U=\frac{kx^2}{2}$

Replacing:

$U=\frac{k(\frac{1}{3}A)^2}{2}\\U=\frac{kA^2}{18}$

4 0

3 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

4 years ago

a tire rolls from rest to an angular velocity of 6 radians per second. the angle it turns through is 12 radians. what is the ang

True [87]

Answer:

$\alpha =1.5\ rad/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega_i=0$

Final angular speed, $\omega_f=6\ rad/s$

Angular displacement, $\theta=12\ rad$

We need to find the angular acceleration of the tire. We can find it using the third equation of rotational kinematics. So,

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{6^2-0^2}{2\times 12}\\\\\alpha =1.5\ rad/s^2$