What are the two opposing forces that keep stars between collapsing and inflating?

Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.

skelet666 [1.2K]

Answer:

Assuming b = 9.3i + 9.5j (b = 931 + 9.5 is wrong):

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

Assuming b = 9.3i + 9.5j (b = 931 + 9.5 is wrong) we can proceed as follows:

a) The vectorial product, a×b is:

$a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k$

b) The escalar product a·b is:

$a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43$

c) Asumming (a + b)·b instead a+b·b we have:

$(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17$

d) The component of a along the direction of b is:

$a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66$

I hope it helps you!

5 0

3 years ago

) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in

Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

$T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}$

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

4 0

3 years ago

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown

ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

$P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,$