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erastova [34]
3 years ago
12

Suppose you saw a waxing gibbous moon. What phase would you expect one week later?

Physics
1 answer:
sattari [20]3 years ago
3 0

The gibbous moon spends roughly a week waxing, from the First Quarter until the Full Moon.  (This is the second week following a New Moon.)

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True / False
devlian [24]
1. True
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2 years ago
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A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r
katen-ka-za [31]

Answer:

  • k = 167.33 N/m

Explanation:

  • The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
  • v = c = λ*f

        where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =  

        4.92*10⁷ m.

        Solving for f, we get the frequency of the radio waves:

        f = 6.1 Hz

  • Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by  a fixed relationship between the spring constant k and the mass m, as follows:

       \omega_{o}^{2} =\frac{k}{m}  (1)

  • Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

       \omega = 2*\pi *f (2)    

  • We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
  • k = 4*\pi ^{2}*f^{2} *m = 4*\pi ^{2} * (6.1Hz)^{2} * 0.114 kg = 167.33 N/m

3 0
3 years ago
Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the
nydimaria [60]

Answer:F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

F_1=\frac{kq(-q)}{(L\sqrt{2})^2}

where  L\sqrt{2}=Distance between the two charges

F_1=-\frac{kq^2}{2L^2}

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

F_2=\frac{kq(-q)}{(L)^2}

The magnitude of force by both the  charge is same but at an angle of 90^{\circ}

thus combination of two forces at 2 and 3 will be

F'=\sqrt{2}\frac{kq^2}{2L^2}

Now it will add with force due to 1 charge

Thus net force will be

F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

6 0
3 years ago
The base of a pyramid covers an area of 13.0 acres (1 acre = 43,560 ft2) and has a height of 481 ft. If the volume of a pyramid
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V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3
3 0
2 years ago
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What is also known as watered carbons
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The name carbohydrate means "watered carbon" or carbon with attached water molecules. Many carbohydrates have empirical formuli which would imply about equal numbers of carbon and water molecules. For example, the glucose formula C6H12O6 suggests six carbon atoms and six water molecules.
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2 years ago
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