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svet-max [94.6K]

3 years ago

10

A wave has a wavelength of 20 mm and a frequency of 5 Hz what is the speed?

1 answer:

yKpoI14uk [10]3 years ago

7 0

The answer is 100mm/s. I hope this helps :)

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An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000

Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷

(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

= 2.0884 x 10⁻⁴ f²

| Z | = √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ] = 1.6 x 10⁵

   (1.6 x 10⁵)² = (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

   (2.56 x 10¹⁰) - (4.624 x 10⁷) = 2.0884 x 10⁻⁴ f²

Frequency² = (2.56 x 10¹⁰ - 4.624 x 10⁷) / 2.0884 x 10⁻⁴

      = 2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

=    1.224 x 10¹⁴

      =      122,400 GHz    <== my calculation

    11.1 MHz <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation.

7 0

3 years ago

When an object of weight w is suspended from the center of a massless string?

igomit [66]

I attached a picture of the diagram associated with this question.

Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ

The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ

4 0

3 years ago

Read 2 more answers

How do sinkholes occur? Overburden falls into the space left behind when bedrock is dissolved. Bedrock falls under the pressure

BabaBlast [244]

Answer: over burden is dissolved by water wind and acids

6 0

3 years ago

Astronauts on the first trip to Mars take along a pendulumthat has a period on earth of 1.50 {\rm s}. The period on Mars turns o

svetlana [45]

Answer:

3.7 m/s^2

Explanation:

The period of a simple pendulum is given by:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum and g is the free-fall acceleration on the planet.

Calling L the length of the pendulum, we know that:

$T_e = 2 \pi \sqrt{\frac{L}{g_e}}=1.50 s$ is the period of the pendulum on Earth, and $g_e = 9.8 m/s^2$ is the free-fall acceleration on Earth

$T_m = 2 \pi \sqrt{\frac{L}{g_m}}=2.45 s$ is the period of the pendulum on Mars, and $g_m = ?$ is the free-fall acceleration on Mars

Dividing the two expressions we get

$\frac{T_e}{T_m}=\sqrt{\frac{g_m}{g_e}}$

And re-arranging it we can find the value of the free-fall acceleration on Mars:

$g_m = g_e \frac{T_e^2}{T_m^2}=(9.8 m/s^2)\frac{(1.50 s)^2}{(2.45 s)^2}=3.7 m/s^2$

4 0

3 years ago

Which image is observe by seeing a distant thing using magnifying glass?

lesya692 [45]

Answer:

1/f = 1/o + 1 /i thin lens equation

1 / i = 1/f - 1/o = (o - f) / (o f) rearranging

or i = o f / (o - f) for the image distance

This image will be positive (real) with values given

m = -i / o magnification

m = - f / (o - f) shows m is negative and the image inverted

Also, since o here is large m will be small and the image is small

I think they mean A as the right answer

Different texts use different terms but here i is image distance and o the object distance

7 0

3 years ago

Read 2 more answers