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3 years ago

The higher an electromagnetic wave's frequency, the lower its

Physics

2 answers:

Pepsi [2]3 years ago

7 0

Answer:

The higher an electromagnetic wave's frequency, the lower its wavelength

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tensa zangetsu [6.8K]3 years ago

3 0

Answer:

The higher an electromagnetic wave's frequency, the lower its wavelength.

Explanation:

The frequency of a electromagnetic wave is given by the following relation :

$c=f\times \lambda$

Where

c = speed of light

f = frequency of a wave

$\lambda$ = wavelength

$f=\dfrac{c}{\lambda}$

It is clear that the frequency of a wave has inverse relationship with its wavelength. So, higher the electromagnetic wave's frequency, the lower its wavelength.

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A sphere of radius 5.15 cm and uniform surface charge density +12.1 µC/m2 exerts an electrostatic force of magnitude 35.9 ✕ 10-3

rosijanka [135]

The radius of the sphere is r=5.15 cm=0.0515 m, and its surface is given by
$A=4 \pi r^2 = 4 \pi (0.0515 m)^2 = 0.033 m^2$

So the total charge on the surface of the sphere is, using the charge density
$\rho=+1.21 \mu C/m^2 = +1.21 \cdot 10^{-6} C/m^2$ :
$Q= \rho A = (+1.21 \cdot 10^{-6} C/m^2)(0.033 m^2)=4.03 \cdot 10^{-8}C$

The electrostatic force between the sphere and the point charge is:
$F=k_e \frac{Qq}{r^2}$
where
ke is the Coulomb's constant
Q is the charge on the sphere
$q=+1.75 \muC = +1.75 \cdot 10^{-6}C$ is the point charge
r is their separation

Re-arranging the equation, we can find the separation between the sphere and the point charge:
$r=\sqrt{ \frac{k_e Q q}{F} }= \sqrt{ \frac{(8.99 \cdot 10^9 Nm^2 C^{-2})(4.03 \cdot 10^{-8} C)(1.75 \cdot 10^{-6}C)}{35.9 \cdot 10^{-3}N} }=0.133 m=13.3 cm$

8 0

3 years ago

A child pulls a wagon at a constant velocity of 4.0 m/s for 4.0 minutes along a level sidewalk. The child does this applying a 2

natita [175]

Answer:

288kj

Explanation:

3 0

3 years ago

A 2.4-kg cart is rolling along a frictionless, horizontal track towards a 1.7-kg cart that is held initially at rest. The carts

inessss [21]

Answer:

Explanation:

Mass of first cart M1=2.4kg

Velocity of first cart U1=4.1m/s

Mass of second cart M2=1.7kg

Second cart is initially at rest U2=0

After an instant, the velocity of the second cart is U2=-2.8m/s

Now after collision the two cart move together with the same velocity I.e inelastic collision

Using conservation of momentum

Momentum before collision, = momentum after collision

M1U1 + M2U2 = (M1+M2)V

2.4×4.1 + 1.7× -2.8 =(2.4+1.7)V

9.84 - 4.76 = 4.1V

5.08=4.1V

V=5.08/4.1

V=1.24m/s

The momentum of the two cart at that instant is

M1U1+M2U2

2.4×4.1 + 1.7× -2.8

9.84 - 4.76

5.08kgm/s

So the momentum at the instant the velocity is 4.1m/s for cart 1 and -2.8m/s for cart 2 is 5.08kgm/s

3 0

3 years ago

Earthquakes with a very deep focus are usually located along ________.

steposvetlana [31]

Earth quakes are commonly found along fault lines or tectonic plates.

Hope this helped :-) <span />

4 0

3 years ago

Read 2 more answers

An object has a kinetic energy of 14 J and a mass of 17 kg , how fast is the object<br> moving?

Vlad1618 [11]

Use the formula:

E = (1/2) * m * v ^2

For unit consistency, make sure that:

E energy is in joules (J)
m mass is in kilograms (kg)
v velocity is in meters per second (m/s)
So, substituting to the equation:

14= (1/2) * (17) * v^2

By algebraic manipulation, we derive:

v = (2*14/17)^0.5

v = 1.2834 m/s
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