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GREYUIT [131]
3 years ago
5

Help Me Fast Please!!

Physics
1 answer:
rusak2 [61]3 years ago
5 0

Answer:

There is only a beat frequency when the frequency heard (perhaps a tuning fork) is different from the frequency from  the object in question (piano string, etc.).

The beat frequency equals the difference of the frequencies of the sources.

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Sunglasses that reduce glare take advantage of which kind of wave
marysya [2.9K]

Answer:

It should be option B polarization

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3 years ago
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Where does the majority of earth's water reside?
tensa zangetsu [6.8K]
Depends on which, most pf water in general resides within the ocean but fresh water resides mostly within ice caps in the arctic, Hope this helps
~Marquest
6 0
3 years ago
A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh
zheka24 [161]

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

F=ILB\ sin\theta

F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

7 0
4 years ago
Read 2 more answers
A car travelled a distance of 200 m with initial velocity of 216 km/hr. Calculate the acceleration
pogonyaev

Answer:

a = 16\ m/s^2

Explanation:

When the velocity changes uniformly, the object has a constant acceleration. The acceleration, the velocities, and the distance are related by the equation:

v_f^2=v_o^2+2ax

Where:

vf = final velocity

vo = initial velocity

a = acceleration

x = distance

Solving for a:

\displaystyle a=\frac{v_f^2-v_o^2}{2x}

The car travels a distance of x=200 m and the velocities are:

vo = 216 Km/h

vf = 360 Km/h

Both velocities must be converted to meters by seconds.

vo = 216 Km/h *1000/3600 = 60 m/s

vf = 360 Km/h *1000/3600 = 100 m/s

The acceleration is:

\displaystyle a=\frac{100^2-60^2}{2*200}

\displaystyle a=\frac{10000-3600}{400}

\displaystyle a=\frac{6400}{400}

\mathbf{a = 16\ m/s^2}

7 0
3 years ago
A 6 ft tall person walks away from a 10 ft lamppost at a constant rate of 5 ft/s. What is the rate (in ft/s) that the tip of the
nalin [4]

Answer:

12.5 ft/s

Explanation:

Height of person = 6 ft

height of lamp post = 10 ft

According to the question,

dx / dt = 5 ft/s

Let the rate of tip of the shadow moves away is dy/dt.

According to the diagram

10 / y = 6 / (y - x)

10 y - 10 x = 6 y

y = 2.5 x

Differentiate both sides with respect to t.

dy / dt = 2.5 dx / dt

dy / dt = 2.5 (5) = 12.5 ft /s

8 0
4 years ago
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