The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.
Let:
b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°
We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.
Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.
In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
For a standing wave on a string, the wavelength is equal to twice the length of the string:
In our problem, L=50.0 cm=0.50 m, therefore the wavelength of the wave is
And the speed of the wave is given by the product between the frequency and the wavelength of the wave:
<u>Answer:</u>
The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".
<u>Explanation:</u>
The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.
The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.
Answer:
The density of gold is of 18 grams per cm3.
Explanation:
The mass density of a homogeneous material expresses how much mass of that material is present in a given volume. Since the density of an object is obtained by dividing its mass by its volume, to obtain the density of gold, its 90 grams of mass must be divided by its 5 cm3 volume, performing the following calculation:
90/5 = X
18 = X
Thus, the density of gold is 18 grams per cm3.
