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barxatty [35]
4 years ago
9

What are some examples of constructive interference and the Doppler Effect in real life​

Physics
1 answer:
trapecia [35]4 years ago
8 0

Answer there really are not any examples to use

Explanation:

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Sloan [31]
WoAh that’s kind bro
5 0
3 years ago
A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is
kenny6666 [7]

Answer:

f2/f1 = \sqrt{2}

Explanation:

From frequency of oscillation

f = 1/2pi *\sqrt{k/m}

Initially with the suspended string, the above equation is correct for the relation, hence

f1 = 1/2pi *\sqrt{k/m}

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

f2 = 1/2pi *\sqrt{2k/m}

Employing f2/ f1, we have

f2/f1 = \sqrt{2}

3 0
3 years ago
Which of the following statements are not true regarding sound?
Likurg_2 [28]

Answer:

All  are  not true.

Explanation:

Sound needs medium to travel.

sound speed is fastest in solids.

5 0
3 years ago
Mountain pull. A large mountain can slightly affect the direction of ""down"" as determined by a plumb line. Assume that we can
Nastasia [14]

Answer:

6.18 um

Explanation:

The plumb line will be pulled down by a combination of the gravitationall pull of Earth and of the mountain. The Earth pulls down and the mountain to the side. Because of this it will fall not in a straight line down, but slightly to the side. Since the plumb line will follow the compound gravity we can imagine a rectangle triangle formed by the plumb line, a vertical line that ends at the same height as the plumb line, and the sideways displacement.

The total gravity will be proportional to the plumb line lenght, the vertical line will be proportional to Earth's gravity and the sideways displacement to the mountain pull.

The gravity of Earth is 9.81 m/s^2

The pull of the mountain will be defined by Newton's law of universal gravitation:

F = G \frac{m1 * m2}{r^2}

Where

F: pull force

G: universal gravitational constant (6.67e-11 m^3/(kg * s)

m1: mass of the mountain

m2: mass of the plumb

r: distance between mountain and plumb (3 km in this case)

If we divide both sides by m2 we obtain the acceleration towards the mountain of the plumb

a = G \frac{m1}{r^2}

Now we need the mass of the mountain. This will be its volume times it's density. The volume depends on the radius (since we consider it as a sphere)

m1 = \delta * \frac{4}{3} * \pi * r^3

m1 = 2.6e3 * \frac{4}{3} * \pi * 1000^3 = 1.09e13 kg

So, the acceleration on the plumb will be

a = 6.67e-11 \frac{1.09e13}{3000^2} = 8.08e-5 m/s^2

This is very small compared to the pull of Earth, so we can make an approximation that the length of the plumb line is equal to vertical line.

We can use the principle of similar triangles to say that:

\frac{\Delta x}{L} = \frac{a}{g}

So:

\Delta x = \frac{a * L}{g} = \frac{8.08e-5 * 0.75}{9.81} = 6.18e-6 m = 6.18 um

3 0
3 years ago
If Superman wants to slow down a fast car with speed 20 m/s and mass 1000 kg, how much force in N does he need to apply if he wa
saveliy_v [14]

Answer:

Force of 6250N is required to slow down the body to zero.

Explanation:

Speed(V)=20m/s

Initial speed (u) =0m/s

Mass(M)=1000kg

Time(t)=3.2seconds

Using first equation of motion to find the acceleration of the body:

V=U + at

20m/s=0 + a*3.2s

Therefore, a=(20m/s)/3.2s

a=6.25m/s^2

To find the force required to stop the body or get it down to zero:

Force(F)=mass x acceleration

F=(1000kg)*(6.25m/s^2)

F=6250N.

Therefore a force of 6250N is required to slow it down.

8 0
3 years ago
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