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3 years ago

What are the characteristics of low energy waves?

2 answers:

8 0

I think the correct answer from the choices presented above is the first option. The characteristics of low energy waves are long wavelengths and low frequencies. Energy is established to be indirectly proportional to wavelengths and frequencies.

gregori [183]3 years ago

6 0

The correct answer to the question is : A) Long wavelengths and low frequencies.

EXPLANATION:

From Planck's quantum theory we know that the energy of any electromagnetic wave having certain frequency is E = $h\nu$ .

Here, $\nu$ is the frequency of the wave.

Hence, more is the frequency of the wave, the wave is more energetic.

We know that wavelength and frequency are inversely proportional to each other.

Hence, a low energy wave means a wave having long wavelength or low frequency.

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Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

8 0

2 years ago

Tarzan (who has mass 80.0 kg) is running across the jungle floor with speed 7.00 m/s as

Lesechka [4]

We are given:

Mass of Tarzan before swinging = 80 kg

Mass of Tarzan when swinging = 80 + 15 = 95 kg

Velocity of Tarzan = 7 m/s

The height of the rock Tarzan's monkey is sitting on = 3 m

__________________________________________________________

Momentum of Tarzan before swinging:

We know that:

Momentum = Mass*Velocity

Momentum = 80 * 7

Momentum = 560 kg m/s

__________________________________________________________

Speed of Tarzan after grabbing the bananas:

The momentum of Tarzan will remain the same but his mass will increase

So, Since Momentum = New Mass* velocity

560 = 95 * v [where v is the velocity of Tarzan]

v = 5.9 m/s

__________________________________________________________

<h3>Finding the Initial and Final KE and PE:</h3>

Here, KE = Kinetic Energy and PE = Potential Energy

Initial and Final KE:

We know that KE = 1/2*(mv²)

Initial KE:

Initial KE = 1/2*(mv²) [where v is the velocity after picking the bananas]

Initial KE = 1/2*(95*5.9²)

Initial KE = 1653.5 Joules

Final KE:

Final KE = 1/2*(mv²)

[where v is the velocity at the maximum point of the swing]

Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s

Final KE = 1/2*(95*0²)

Final KE = 0 Joules

Initial and Final PE:

We know that:

PE = mgh

[where g is the acceleration due to gravity and h is the height]

Initial PE:

Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0

Initial PE = 95*10*0

Initial PE = 0 Joules

Final PE:

Let the maximum height of Tarzan be h m

Final PE = 95*10*h

Final PE = 950(h)

__________________________________________________________

<h3>Finding the maximum height Tarzan will reach:</h3>

Here, KE = Kinetic Energy and PE = Potential Energy

From the law of conservation of momentum, we know that:

Initial KE + Initial PE = Final KE + Final PE

Replacing the variables:

1653.5 + 0 = 0 + 950h

1653.5 = 950h

h = 1653.5/950 [dividing both sides by 950]

h = 1.74 m

Therefore, the maximum height reached by Tarzan is 1.74 m

but since his monkey is sitting 3 m high, he will NOT be able to reach his monkey

6 0

2 years ago

A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it

asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0

2 years ago

the royal Gorge Bridge in Colorado rises 321 m above the Arkansas river. suppose you kick a rock horizontally off the bridge. Th

KengaRu [80]

Answer:

2.48 m/s

Explanation:

We can use the kinematic equation,

s = ut +½at²

Where

s = displacement

u = initial velocity

t = time taken

a = acceleration

Using the equation in vertical direction,

321 = 0×t +½×g×t², u = 0 because initial vertical velocity is 0

We get t = 8.01 s

Using the equation in the horizontal direction,

52 = u×8.01 +½×0×(8.01)²,. a = 0 because no unbalanced force act on object in that direction

So u = 2.48 m/s

5 0

3 years ago

What are the units for gravitational field strength ?

Jobisdone [24]

Answer:

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

7 0

3 years ago

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