The Doppler Effect for sound occurs when:

Evan drew a diagram to illustrate radiation.

Vesna [10]

The correct answer is:

D. Electromagnetic waves.

The arrows represent electromagnetic waves.

$|Huntrw6|$

8 0

3 years ago

Two people carry a heavy electric motor by placing it on a light board 2.45 m long. One person lifts at one end with a force of

maxonik [38]

Answer:

W=1055N

Explanation:

In order to solve this problem, we must first do a drawing of the situation so we can visualize theh problem better. (See attached picture)

In this problem, we will ignore the board's weight. As we can see in the free body diagram of the board, there are only three forces acting on the system and we can say the system is in vertical equilibrium, so from this we can say that:

$\sum F=0$

so we can do the sum now:

$F_{1}+F_{2}-W=0$

when solving for the Weight W, we get:

$W=F_{1}+F_{2}$

and now we can substitute the given data, so we get:

W=410N+645N

W=1055N

5 0

3 years ago

Please help I'm confused on where to start and with the graph.

tankabanditka [31]

I don't think anyone knows the answer

3 0

3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

$p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}$

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

$F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N$

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

$F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}$

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0

3 years ago

A beam of light in air is incident at an angle of 30º to the surface of a rectangular block of clear plastic (n = 1.46). The lig

Aneli [31]

Answer:

θ = 30°

Explanation:

Firts, the angle when the beam of light passes through the block cam be calculated using Snell Law:

$n_{1}sin(\theta_{1}) = n_{2}sin(\theeta_{2})$

<u>Where</u>:

n₁: is the index of refraction of the incident medium (air) = 1

θ₁: is the incident angle = 30°

n₂: is the medium 2 (plastic) = 1.46

θ₂: is the transmission angle

Hence, θ₂ is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1*sin(30)}{1.46} = 0.34 \rightarrow \theta_{2} = 20.03 ^{\circ}$

Now, when the beam of light re-emerges from the opposite side, we have:

n₁: is the index of refraction of the incident medium (plastic) = 1.46

θ₁: is the incident angle = 20.03°

n₂: is the medium 2 (air) = 1

θ₂: is the transmission angle

Hence, the angle to the normal to that surface (θ₂) is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1.46*sin(20.03)}{1} = 0.50 \rightarrow \theta_{2} = 30 ^{\circ}$

Therefore, we have that the beam of light will come out at the same angle of when it went in, since, it goes from air and enters to a plastic medium and then enters again in this medium to go out to air again. This was proved using the Snell Law.

I hope it helps you!

5 0

2 years ago