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3 years ago

10

If a wave has amplitude of 2 meters, a wavelength of 2 meters, and a frequency of 10 Hz, and a period of 1 second, then at what

speed is the wave moving?

1 answer:

Serhud [2]3 years ago

6 0

Answer:

20 m/s

Explanation:

The speed of a wave is given by:

$v=\lambda f$

where

$\lambda$ is the wavelength

f is the frequency

v is the speed

For the wave in this problem,

f = 10 Hz is the frequency

$\lambda=2 m$ is the wavelength

So the speed is

$v=(10 Hz)(2 m)=20 m/s$

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Worth 25 points on my exam

I am Lyosha [343]

Initial velocity=20m/s
Final velocity=0m/s(As the car stops)
Acceleration=-8m/s^2
Distance=s=26m

We need to verify the thrid equation of kinematics here

$\\ \tt\longmapsto v^2-u^2=2as$

$\\ \tt\longmapsto 20^2=2(-8)s$

$\\ \tt\longmapsto 400=-16s$

$\\ \tt\longmapsto s=|400/-16|$

$\\ \tt\longmapsto s=25m$

The squirrel has a good luck ,Car gets stopped just 1m away from the squirrel .

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2 years ago

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Why are collisions so important to physicists

Komok [63]

They study the causes

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3 years ago

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

A

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

B

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

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3 years ago

If the potential due to a point charge is 490 V at a distance of 10 m, what are the sign and magnitude of the charge?

uysha [10]

Answer:

+5.4×10⁻⁷ C

Explanation:

Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)

The formula for potential is

V = kq/r............................ Equation 1

Where V = electric potential, k = proportionality constant, q = charge, r = distance.

making q the subject of the equation,

q = Vr/k............................ Equation 2

Given: V = 490 V, r = 10 m,

Constant: k = 9×10⁹ Nm²/C²

Substitute into equation 2

q = 490(10)/(9×10⁹)

q = 5.4×10⁻⁷ C

q = +5.4×10⁻⁷ C

Hence the charge is +5.4×10⁻⁷ C

7 0

3 years ago

If an object starts from rest, what is its initial velocity?

Vladimir79 [104]

Answer:

0 m/s

Explanation:

velocity= change in displacement/ time

at rest, the ball does not travel any distance

0/ t

=0

4 0

3 years ago