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3 years ago

Which three statements describe Electromagnetic waves

Physics

1 answer:

Brums [2.3K]3 years ago

7 0

Answer:

As you learned in Wave Motion, all waves have amplitude, wavelength, velocity and frequency.hope that helps

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A coin rests on a record 0.13 m from its center. The record turns on a turntable that rotates at variable speed. The coefficient

igor_vitrenko [27]

Answer:

Maximum speed of the coin, v = 0.607 m/s

Explanation:

It is given that,

A coin rests on a record 0.13 m from its center i.e. the radius of the circular path, r = 0.13 m

The coefficient of static friction between the coin and the record is, $\mu=0.29$

The centripetal force acting on the coin is balanced by the frictional force. Its mathematical relation is given by :

$\dfrac{mv^2}{r}=\mu mg$

$v=\sqrt{\mu gr}$

$v=\sqrt{0.29\times 9.8\times 0.13}$

v = 0.607 m/s

So, the maximum coin speed at which it does not slip is 0.607 m/s. Hence, this is the required solution.

3 0

3 years ago

A new planet is discovered orbiting a distant star. Observations have confirmed that the planet has a circular orbit with a radi

Mariulka [41]

Answer:

Mass of star is $1.31\times10^{35}$ kg.

Explanation:

The cube of orbital radius is equal to the square of its orbital time period is known as Kepler's law.

$T^{2} = (\frac{4\pi^{2} }{GM})r^{3}$ .....(1)

Here T is time period, r is orbital radius, G is universal gravitational constant and M is the mass of the star.

According to the problem,

Time period, T = 109 days = 109 x 24 x 60 x 60 s = 9.41 x 10⁶ s

Orbital radius, r = 18 AU = 18 x 1.496 x 10¹¹ m = 2.70 x 10¹² m

Gravitational constant, G = 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻²

Substitute these values in equation (1).

$(9.41\times10^{6}) ^{2} = (\frac{4\pi^{2} }{6.67\times10^{-11}\times M})(2.70\times10^{12}) ^{3}$

M = $1.31\times10^{35}$ kg

3 0

3 years ago

Q. Calculate the annual energy

melomori [17]

Answer:

60 kWh

Explanation:

The computation of the annual energy consumption in KW-h is shown below:

As we know that

1 kw = 1000 w

So, for 1400 W it would be

= 1,400 ÷ 1,000

= 1.4 kW

Now the number of hours it used in a year

= 7 minutes × 365 days ÷ 60 minutes

= 42.58333 hours

So in one year it used

= 1.4 kW × 42.58333

= 59.61 kWh

= 60 kWh

3 0

3 years ago

In 1991 at smith college, in massachusetts, ferdie adoboe ran 1.00 × 102 m backward in 13.6 s. suppose it takes adoboe 2.00 s to

Alik [6]

as it is given that it covers a total distance 1 * 10^2 m

total time taken by it = 13.6 s

now the average speed is given as ratio of total distance and total time

$v = \frac{d}{t}$

$v = \frac{1* 10^2 }{13.6}$

$v = 7.35 m/s$

so the average speed will be 7.35 m/s

now if it starts from rest and achieve the final speed as 7.35 m/s

now we can use kinematics

$v_f = v_i + at$

$7.35 = 0 + a* 2$

$a = 3.68 m/s^2$

so its acceleration will be 3.68 m/s^2

4 0

4 years ago

What are the differences between the practical and the ideal pendulum

ankoles [38]

lf a heavy point mass is suspended by a weightless, inextensible and perfectly flexible string from a rigid support, then this arrangement is called simple pendulum.

In practice, however, these requirements cannot be fulfilled. So we use a practical pendulum.

A practical pendulum consists of a small metallic solid sphere suspended by a fine silk thread from a rigid support. This is the practical simple pendulum which is nearest to the ideal simple pendulum.

Note :

The metallic sphere is called the bob.

When the bob is displaced slightly to one side from its mean position and released, it oscillates about its mean position in a vertical plane.

4 0

3 years ago