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rewona [7]
3 years ago
13

The magnetic field of a light wave oscillates parallel to a y axis and is given by by = bmsin(kz – ?t. (a in what direction does

the wave travel?
Physics
1 answer:
-Dominant- [34]3 years ago
6 0
In this case the coordinate z is multiplied by your wave vector k, so it travels along z.  Note that the electric field will be oscillating along the x direction.  H and E are perpendicular to the direction of travel in vacuum.
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Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq
vekshin1

Answer:

Time period, T=1.11\times 10^{-8}\ s

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

It is required to find the period of the wave.

The reciprocal of frequency is called time period of a wave. It can be given by :

T=\dfrac{1}{f}\\\\T=\dfrac{1}{89.3\times 10^{6}\ Hz}\\\\T=1.11\times 10^{-8}\ s

So, the period of the wave is 1.11\times 10^{-8}\ s.

5 0
3 years ago
A car with a mass of 3 Kg and velocity of 40 m/s collided with a truck of a velocity of 60 m/s, if the momentum is conserved wha
Luden [163]

Answer:

the mass of the truck is 2 kg.

Explanation:

Given;

mass of the car, m₁ = 3 kg

initial velocity of the car, u₁ = 40 m/s

initial velocity of the truck, u₂ = 60 m/s

let the mass of the truck = m₂

Apply the principle of conservation of linear momemtum;

m₁u₁ = m₂u₂

m₂ = (m₁u₁) / u₂

m₂ = (3 x 40) / (60)

m₂ = 2 kg

Therefore, the mass of the truck is 2 kg.

4 0
3 years ago
A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa
kotykmax [81]
Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is v_i=180~m/s, the final velocity is v_i=0~m/s, and the total time of the motion is \Delta t=0.02~s, so the acceleration is given by
a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2 
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m
So, the bullet penetrates the sandbag 1.8 meters.
5 0
3 years ago
If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity lev
nekit [7.7K]

Answer:

(c) 97 dB sound intensity level

Explanation:

We have given the intensity of the loud car horn I=0.005w/m^2

We know that I_O=10^{-12}w/m^2

Now the sound intensity level is given by \beta =10log\frac{I}{I_0}=10log\frac{0.005}{10^{-12}}=96.98dB , which is nearly equal to 97

So the sound intensity level will be 97 dB

So option (c) will be the correct option

4 0
3 years ago
on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak
sineoko [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>  

That's about 2.9 G's .  Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't.  Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

Force on the airplane = 504,000 Newtons

That's about 113,000 pounds !  Maybe the part of the airplane that the catapult pushes on can't handle any more force than that.  Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds.  Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes !  Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

8 0
3 years ago
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