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2 years ago

What best describes the angle between a changing electric field and the electromagnetic wave produced by it?

Physics

1 answer:

mel-nik [20]2 years ago

3 0

The correct answer here is Option A, always equal to a right angle.

Keep in mind that the phase difference that exists between the electric field and the magnetic wave is 90 °, this means that the changing electric field and the electromagnetic wave are always perpendicular and will be forming a right angle.

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Need help on this thank you

Semmy [17]

Answer:

TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.

7 0

2 years ago

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±

Alla [95]

Answer:

P = 1 (14,045 ± 0.03 ) k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

Δ (Pₓ / Py) =?

Let's start by finding the momentum of each vehicle

car X

Pₓ = m vₓ

Pₓ = 2.34 2.5

Pₓ = 5.85 kg m

car Y

Py = 2,561 3.2

Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

ΔPₓ = m Δv + v Δm

ΔPₓ = 2.34 0.01 + 2.561 0.01

ΔPₓ = 0.05 kg m

Δ $P_{y}$ = m Δv + v Δm

ΔP_{y} = 2,561 0.01+ 3.2 0.001

ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

P = Pₓ / $P_{y}$

ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

ΔP = 0.006 + 0.0026

ΔP = 0.009 kg m

The result is

P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s

7 0

2 years ago

A transformer has input voltage and current of 12 V and 4 A

Advocard [28]

Explanation:

It is given that,

Input voltage, $V_i=12\ V$

Input current, $I_i=4\ A$

Output current, $I_o=0.8\ A$

Number of turns in the secondary side of transformer, $N_s=1177$

We need to find the number of turns in the primary side of the transformer. The current to the number of turns in the input and output is given by :

$\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}$

Substituting all the above values

So,

$N_p=\dfrac{N_sI_s}{I_p}\\\\N_p=\dfrac{1177\times 4}{0.8}\\\\N_p=5885$

So, the number of turns in primary side of the transformer is 5885.

4 0

2 years ago

An 1100kg car reached 30m/s in 9s from rest. What force did the engine provide?

malfutka [58]

Answer:

≅3666.67 N

Explanation:

Use Newton's 2nd law, F = ma where F=force applied, m = mass of the object,

a = acceleration acquired by the object.

a= (v-u)/t where v = final velocity, u = initial velocity and t = time taken

calculate a = (30-0)/9 ≅ 3.33 m/s2

Then F = 1100×a = 3666.67 N

3 0

2 years ago

A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin

JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

$f_s = f(\dfrac{v}{v-(-v_s)})$

$f_s = f(\dfrac{v}{v+v_s})$

$v_s = v[\dfrac{f_s}{f_o}-1]$

$= 343[\dfrac{508}{490}-1]$

$v_s=12.6 m/s$

distance the tunning fork has fallen

$y_1=\dfrac{v^2}{2a_y}$

$=\dfrac{12.6^2}{2\times 9.8}$

=8.1 m

now, time required for the observed will be

$t = \dfrac{8.1}{343} = 0.023 s$

now, for the distance calculation

$y_2 = u\ t + \dfrac{1}{2}at^2$

$= 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2$

=0.293 m

total distance

= 8.1 + 0.293 = 8.392 m

3 0

2 years ago