All categories

3 years ago

Can we have the Doppler effect with light waves? How?

Physics

1 answer:

Mekhanik [1.2K]3 years ago

7 0

Answer:

Besides sound and radio waves, the Doppler effect also affects the light emitted by other bodies in space. If a body in space is "blue shifted," its light waves are compacted and it is coming towards us. If it is "red shifted" the light waves are spread apart, and it is traveling away from us.

Explanation:

You might be interested in

A 210Ω resistor is connected in a circuit with a 110V battery. What total amount of charge passes through a point in the circuit

irina1246 [14]

Answer: 62.86 coulombs

Explanation:

Resistance (R) = 210Ω

Voltage of battery (V) = 110V

total amount of charge (Q) = ?

Time (T) = 2 minutes

The SI unit of time is seconds so convert 2 minutes to seconds

(If 1 minute = 60 seconds

2 minutes = 2 x 60 = 120 seconds)

To get the total charge, first get the current (I) flowing in the circuit by applying the formula V = IR

110V = I x 210Ω

I = 110V/210Ω

I = 0.524 Amps

Then, apply the formula

Charge = current x time

i.e Q = IT

Q = 0.524 Amps x 120 seconds

Q = 62.86 coulombs

Thus, 62.86 coulombs of charge passes through the circuit.

6 0

4 years ago

The splitting of an atom is called Blank Space __________.

horsena [70]

Fission is the answer

8 0

3 years ago

A nonconducting rod of length L = 8.15 cm has charge –q = -4.23 fC uniformly distributed along its length.(a) What is the linear

vichka [17]

Answer:

a) λ = 5.19 10⁻⁴ C/m , b) E = 1,573 10⁻³ N/C , c) the direction of the field is directed to the bar

Explanation:

a) the linear density defined as the ratio between the charge per unit length

λ = q / l

Let's start by reducing the units to the SI system

L = 8.15 cm (1m / 100cm) = 8.15 10⁻² m

a = 12 cm (1m / 100cm) = 12 10⁻² m

q = -4.23 fC (1 C / 10¹⁵ ft) = -4.23 10⁻¹⁵ C

λ = -4.23 10⁻¹⁵ C / 8.15 10⁻²

λ = 5.19 10⁻⁴ C/m

b) Let's look for the electric field for a point at a distance a from the end of the bar

E = k dq / r²

To simplify the notation, suppose the bar is the x axis. Since the density is constant, we can write it differentially

λ = dq/dx

dq = λ dx

E = k ∫ λ dx / x²

We integrate and evaluate between the lower limits x = a and higher x = L + a. Here we place the test point at the origin of the system

E = k λ (-1 / x)

E = k λ (-1 /(L + a) + 1 /a)

E = k λ (L /a(L + a)

Let's change the density for its value

E = k (q / L) (L / a (L + a)

E = k q 1 /[a(L + a)]

E = 8.99 10⁹ 4.23 10⁻¹⁵ [1 /12 10⁻²(8.15 10⁻² + 12 10⁻²)]

E = 1,573 10⁻³ N/C

c) the direction of the field is directed to the bar, because it has a negative charge

d) now we change the distance a = 50 cm = 0.50 m

Bar

E = 8.99 10⁹ 4.23 10⁻¹⁵ ( 1 /0.5(0.0815 +0.5))

E = 1,308 10⁻⁴ N/C

Charge point

q = -4.23 10⁻¹⁵ C

E = k q / r²

E = 8.99 10⁹ 4.23 10⁻¹⁵ / 0.5²

E = 1.521 10⁻⁴ N/C

7 0

3 years ago

At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m

eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, $t_w$ , is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, $d_w$ , is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking $t_w$ = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × $t_w$ ²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × $t_w$ ²

$t_w$ = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = $t_w$ ≈ 1.392 s

2) The horizontal distance, $d_w$ , in meters from the edge of the pool to where the diver enters the water is given as follows;

$d_w$ = L + v₀ × $t_w$ = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0

3 years ago

A turntable with a moment of inertia of 7.2 × − ⋅ rotates freely with an angular speed of 6.5 ⁄ . Riding on the rim of the turnt

irina1246 [14]

Answer:

turntable with a moment of inertia of 7.2 × − ⋅ rotates freely with an angular speed of 6.5 ⁄ . Riding on the rim of the turntable, 2 from the center, is a hamster. When the hamster walks to the center of the turntable, the angular speed of the turntable becomes ⁄. What is the mass of hamster?

Explanation:

6 0

3 years ago