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3 years ago

The graph indicates what about the relationship between wavelength and frequency

2 answers:

8 0

We can't see the graph.
But is has to show that wavelength and frequency are inversely
proportional. In other words, their product is constant.
That's because their product is the speed of the wave.
If the graph doesn't show that, then the graph is wrong.

Luden [163]3 years ago

3 0

the right answer is A as wavelength decreases, frequency increases

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sergey [27]

Answer:

the pressure due to the water on the diver is 200,000 pascal

pressure = height × density × acceleration due to gravity

p = 20×1000×10

p=200,000 pascal

6 0

2 years ago

A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s

zubka84 [21]

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

K.E(hammer) = ½Mh•Vh²

K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

4 0

3 years ago

What are the units of impulse

attashe74 [19]

Answer:

Newton Second

Explanation:

The SI unit of impulse in Newton Second (N.s)

3 0

2 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$