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3 years ago

6

Is light an electromagnetic wave? why?

1 answer:

djverab [1.8K]3 years ago

8 0

read it carefuly you will understand why?????

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Hi! I have a word bank I need help with please!<br> I have the questions as an attachment

Katena32 [7]

Scott needs to determine the density of a metallic rod. First, he should determine the mass of his sample on the laboratory balance. Second, he should measure the volume of his sample by water displacement. Finally, he can calculate the density by dividing mass/volume.
Hope this helped ;)

4 0

3 years ago

A vector R is resolved into its components, Rx and Ry. If the ratio of is 2, what is the angle that the resultant makes with the

Artemon [7]

The angle is 26.56 degrees

6 0

3 years ago

Read 2 more answers

The tip of the fan blade is 0.61 m from the center of the fan. The fan turns at a constant speed and completes 2 rotation every

Goryan [66]

Answer:

What is the centripetal acceleration of the tip of the fan blade?

6.0 m/s2

48 m/s2

53 m/s2

96 m/s2

Answer is 96

Explanation:

5 0

3 years ago

It is found that the most probable speed of molecules in a gas at equilibrium temperature

kaheart [24]

Answer:

$\frac{T_2}{T_1} = 1$

Explanation:

The root mean square velocity of the gas at an equilibrium temperature is given by the following formula:

$v = \sqrt{\frac{3RT}{M} }$

where,

v = root mean square velocity of molecules:

R = Universal Gas Constant

T = Equilibrium Temperature

M = Molecular Mass of the Gas

Therefore,

For T = T₁ :

$v = \sqrt{\frac{3RT_1}{M} }$

For T = T₂ :

$v = \sqrt{\frac{3RT_2}{M} }$

Since both speeds are given to be equal. Therefore, comparing both equations, we get:

$\sqrt{\frac{3RT_1}{M} }=\sqrt{\frac{3RT_2}{M} }\\\\\frac{T_2}{T_1} = 1$

8 0

3 years ago

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago