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k0ka [10]
4 years ago
10

What is mass loss and how does it figure in the death of a star? is the sun undergoing mass loss

Physics
1 answer:
Debora [2.8K]4 years ago
8 0

111122 is the mass of the star

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The electric forces on both beads are equal.
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Find the lengths of each of the following vectors
Irina18 [472]

Answer:

Explanation:

Generally, length of vector means the magnitude of the vector.

So, given a vector

R = a•i + b•j + c•k

Then, it magnitude can be caused using

|R|= √(a²+b²+c²)

So, applying this to each of the vector given.

(a) 2i + 4j + 3k

The length is

L = √(2²+4²+3²)

L = √(4+16+9)

L = √29

L = 5.385 unit

(b) 5i − 2j + k

Note that k means 1k

The length is

L = √(5²+(-2)²+1²)

Note that, -×- = +

L = √(25+4+1)

L = √30

L = 5.477 unit

(c) 2i − k

Note that, since there is no component j implies that j component is 0

L = 2i + 0j - 1k

The length is

L = √(2²+0²+(-1)²)

L = √(4+0+1)

L = √5

L = 2.236 unit

(d) 5i

Same as above no is j-component and k-component

L = 5i + 0j + 0k

The length is

L = √(5²+0²+0²)

L = √(25+0+0)

L = √25

L = 5 unit

(e) 3i − 2j − k

The length is

L = √(3²+(-2)²+(-1)²)

L = √(9+4+1)

L = √14

L = 3.742 unit

(f) i + j + k

The length is

L = √(1²+1²+1²)

L = √(1+1+1)

L = √3

L = 1.7321 unit

3 0
3 years ago
a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
Delicious77 [7]

Answer:

magnitude=34.45 m

direction=55.52\°

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance d between two points:

d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}} (1)

d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}} (2)

d=\sqrt{1186.81 m^{2}} (3)

d=34.45 m (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

tan \theta=\frac{Y2-Y1}{X2-X1}  (5)

tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}  (6)

tan \theta=\frac{24.8}{19.5}  (7)

Finding \theta:

\theta= tan^{-1}(\frac{24.8}{19.5})  (8)

\theta= 55.52\°  (9) This is the direction of the vector

3 0
3 years ago
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