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frosja888 [35]
3 years ago
13

A similarity between electromagnetic waves and matter waves.

Physics
1 answer:
adell [148]3 years ago
3 0

✦ ✦ ✦ Beep Boop - Blu Bot! At Your Service! Scanning Question . . . Code :

                           Green! Letters and Variables Received! ✦ ✦ ✦

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 Question: : A similarity between electromagnetic waves and matter waves?

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Answer: electromagnetic waves are the combination of electric field and magnetic field and the matter wave related to dicrete particles. They are both waves (obviously), and both are made of particles which obey the equation E = hf. Where E is the energy of a particle and f is its frequency. Matter waves can move at varying speeds, and matter waves can be at rest. EM waves always move at c in vacuum.

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A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?
Nostrana [21]

Answer:

The work done by friction was -4.5\times10^{5}\ J

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

W=\Delta KE

W=K.E_{f}-K.E_{i}

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2

Put the value of m and v

W=0-\dfrac{1}{2}\times1000\times(30)^2

W=-450000
\ J

W=-4.5\times10^{5}\ J

Hence, The work done by friction was -4.5\times10^{5}\ J

3 0
3 years ago
A care package is dropped from an airplane flying at 1000 m above sea level with a velocity of 200 m/s to an island. If it takes
ikadub [295]

Answer:3000

Explanation:

4 0
3 years ago
If a statement is true, select true. if it's false, select false. <br>​
Elena L [17]

3 is false 2 is true and the rest true

7 0
3 years ago
Read 2 more answers
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
Read 2 more answers
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0
3 years ago
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