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ycow [4]
3 years ago
8

The four giant gas planets that are made primarily of lightweight elements are

Physics
1 answer:
Hoochie [10]3 years ago
8 0
D) although recent research (1990's) has shown Uranus and Neptune to be ice giants and have heavier substances.
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Can somebody please help me.
stira [4]

Answer:

what is the image in question

7 0
3 years ago
What is the wavelength of a radio photon from an "am" radio station that broadcasts at 1120 kilohertz?
Alex73 [517]

The wavelength of the radio photon  is 268 m.

Electromagnetic waves travel with the speed of 3×10⁸m/s in vacuum. The speed of a wave <em>c i</em>s related to its frequency<em> f</em> and wavelengthλ as follows:

c=f\lambda

Thus the wavelength is given by,

\lambda =\frac{c}{f}

Substitute 3×10⁸m/s for <em>c</em> and 1120×10³Hz for <em>f</em>.

\lambda =\frac{c}{f}\\ =\frac{3*10^8m/s}{1120*10^3Hz} \\ =267.8 m\\ =268 m

Thus the wavelength of the radio photon from an AM station is 268 m

6 0
3 years ago
If you let him have Czechoslovakia, he is going to.​
valentinak56 [21]

I - ok. Czechoslovakia is officially taken.

4 0
2 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

5 0
3 years ago
A penny is dropped from the top of a building that is 300.0 m tall. Calculate the speed of the penny as it hits the ground. (met
Sauron [17]

We have the equation of motion s = ut + \frac{1}{2} at^2, where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here s = 300 m, u = 0 m/s, a = 9.81 m/s^2

Substituting

   300 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 = 300\\ \\ t =7.82 seconds

Now we have v = u+at, where v is the final velocity

Here u = 0 m/s, a= 9.81 m/s^2 and t = 7.82 seconds

Substituting

     v = 0+9.8*7.82 = 76.68 m/s

The speed with which the penny strikes the ground = 76.68 m/s.

3 0
3 years ago
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