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3 years ago

When a wave is acted upon by an external damping force, what happens to the energy of the wave

Physics

2 answers:

Rasek [7]3 years ago

8 0

The energy of the wave decreases gradually

bazaltina [42]3 years ago

3 0

Answer:

Explanation:

The energy of the wave its determined by the mass of the fuid of the wave and the speed. The amount of motion equation P= m.v

So according to the impulse and amount of motion theorem (ΔP= mvₙ- mv₀).

If the wave is acted upon by an external dumping force, the wave its going to decrease his energy because the reduction of the speed and the mass.

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Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC

Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

$q_1=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 1)

$q_2=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

$F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N$

5 0

3 years ago

A 6 kg tennis ball moves at a velocity of 14 m/s. The ball is struck by a racket, causing it to rebound in the opposite directio

Sergeeva-Olga [200]

Answer and method on photo

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2 years ago

The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

Nadusha1986 [10]

Answer:

The charge resides on the outer surface = $1.245 \times 10^{-12}$ C

Explanation:

Surface area of cell $(A) = 4.3\times 10^{-9} m^{2}$

Separation between two plate $(d) = 1.1 \times 10^{-8} m$

Dielectric constant $(k) = 4.2$

Potential difference $(\Delta V) = 85.7 \times 10^{-3} V$

The capacitance of parallel plate capacitor in free space is given by,

$C = \frac{\epsilon_{o} A }{d}$

Where $\epsilon_{o} =$ permittivity of free space = $8.85 \times 10^{-12}$

The Capacitance of capacitor is increase by $k$ times when it placed in dielectric medium.

$C_{dielectric} = \frac{k \epsilon_{o} A }{d}$

And we know that, $C = \frac{Q}{ \Delta V}$

So charge on the outer surface is given by,

$Q = \frac{k \epsilon A \Delta V }{d}$

$Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3} }{1.1 \times 10^{-8} }$

$Q = 1.245 \times 10^{-12}$

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