Answer:
One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole. The orbital period of A0620-0090 is 7.75hours, the mass of V616 Monocerotis is estimated to be .67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of the orbit of the orange star.
Explanation:
To determine the number of gold bars that would fit in the vault, determine the volume (v) of a single gold bar.
v = l x h x w
where l, h, and w are length, height and width, respectively. Substituting the known values,
v = (6 cm)(2 cm)(3 cm) = 36 cm³
The number of gold bars can be determined by dividing the volume of the vault by the volume of the gold bar,
n = (500 cm³ / 36 cm³) = 13.89
Thus, the maximum number of gold bars in the vault is 13.
Answer:
The combined velocity is 8.61 m/s.
Explanation:
Given that,
The mass of a truck, m = 2800 kg
Initial speed of truck, u = 12 m/s
The mass of a car, m' = 1100 kg
Initial speed of the car, u' = 0
We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.
So, the combined velocity is 8.61 m/s.
A plane mirror forms a virtual image behind the mirror. The image is as far behind the mirror as the object is in front of it. A cannot see his image because the length of the mirror is too short on his side. However, he can see the objects placed at points P and Q, but cannot see the object placed at point R
Hope this helps Buddy!
~ Courtney
Bandwidth is the difference between the upper and lower frequencies in a continuous band of frequencies.
Mathematically can be expressed as,
The upper frequency is 88Hz and the lower frequency is 6, therefore the Bandwidth would be,
The center frequency is given on the basis of the square root of the multiplication of the two reference frequencies, then,
