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xxTIMURxx [149]
3 years ago
11

A particle whose mass is 4 kg moves in xyplane with a constant speed of 2 m/s in the positive x-direction along y = 6 m. Find th

e magnitude of its angular momentum relative to the point (x0, y0), where x0 = 0.9 m and y0 = 10 m. Answer in units of kg m2 /s.
Mathematics
1 answer:
s2008m [1.1K]3 years ago
5 0

Answer: 32.8kgm^2/s

Step-by-step explanation:

1. Angular momentum is a product of linear momentum and the distance of the object from a rotation axis.

2. Linear momentum (L) is given as :

L = mv

Where: m= mass = 4kg

v= velocity= 2m/s

=> L = 8kgm/s

Angular momentum (A) = L x centre distance (d) of object from a rotation axis

d = √(0.9-0)^2 + (10-6)^2 = 4.1m

=> A = 8 x 4.1 = 32.8kgm^2/s.

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Answer:

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Step-by-step explanation:

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2 years ago
What is 863 x 7698? I don’t know the answer.
Flauer [41]
6643374................................................................................
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3 years ago
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find the equation of a circle which passes through the point (2,-2) and (3,4) and whose centre lies on the line x+y=2
Nadusha1986 [10]

Answer:

Equation of the circle

(x - 0.7)² + (y - 1.3)² = 12.58

Step-by-step explanation:

The formula for the equation of a circle is given as:

(x - a)² + (y - b)² = r²,

where(a, b) is the center of the circle and r = radius of the circle.

a) We are told in the question that the equation of the circle passes through point(2, -2)

Hence,

Substituting 2 for x and -2 for y in the equation of the circle.

(x - a)² + (y - b)² = r²

(2 - a)² +(-2 - b)² = r²

Expanding the bracket

(2 - a) (2 - a) + (-2 - b)(-2 - b) = r²

4 - 2a - 2a +a² +4 +2b +2b +b² = r²

4 - 4a + a² + 4 + 4b + b² = r²

a² + b² -4a + 4b + 4 + 4 = r²

a² + b² -4a + 4b + 8 = r²............Equation 1

We are also told that the equation of the circle also passes through point (3,4) also, where 3 = x and 4 = y

Hence,

Substituting 3 for x and 4 for y in the equation of the circle.

(x - a)² + (y - b)² = r²

(3 - a)² +(4 - b)² = r²

Expanding the bracket

(3 - a) (3 - a) + (4 - b)(4- b) = r²

9 - 3a - 3a +a² +16 -4b -4b +b² = r²

9 -6a + a² + 16 -8b + b² = r²

a² + b² -6a -8b + 9 + 16 = r²

a² + b² -6a -8b + 25 = r²..........Equation 2

The next step would be to subtract Equation 1 from Equation 2

a² + b² -4a + 4b + 8 - (a² + b² -6a -8b + 25) = r² - r²

a² + b² -4a + 4b + 8 - a² - b² +6a +8b - -25= r² - r²

Collecting like terms

a² - a² + b² - b² - 4a + 6a + 4b + 8b +8- 25 = 0

2a + 12b -17 = 0

2a + 12b = 17...........Equation 3

Step 2

We are going to have to find the values of a and b in other to get our equation of the circle.

Since the center of the circle(a, b) lies on x + y = 2

Therefore, we have

a + b = 2

a = 2 - b

2a + 12b = 17 ..........Equation 3

Substituting 2 - b for a in

2(2 - b) + 12b = 17

4 - 2b + 12b = 17

4 + 10b = 17

10b = 17 - 4

10b = 13

b = 13/10

b = 1.3

Substituting 1.3 for b in

a + b = 2

a + 1.3 = 2

a = 2 - 1.3

a = 0.7

hence, a = 0.7, b = 1.3

Step 3

We have to find the value of r using points (2, -2)

(x - a)² + (y - b)² = r²

Where x = 2 and y = -2

(-2 - 0.7)² + (-2 - 1.3)² = r²

(-2.7)² + (-3.3)² = r²

1.69 + 10.89 = r²

r² = 12.58

r = √12.58 = 3.55

Step 4

The formula for the equation of a circle is given as:

(x - a)² + (y - b)² = r²,

where(a, b) is the center of the circle and r = radius of the circle

a = 0.7

b = 1.3

r² = 12.58

Equation of the circle =

(x - 0.7)² + (y - 1.3)² = 12.58

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Answer:

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x = 4

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<em>substitute </em>

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Marysya12 [62]

Answer: 0. 327

Step-by-step explanation:

I just used a demos calculator I suggest using demos scientific calculator.

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3 years ago
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