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3 years ago

Find the exact angular velocity in radians per hour wind W= 4.1 rpm

Mathematics

1 answer:

nataly862011 [7]3 years ago

6 0

Answer:

1545.66 rad/h

Step-by-step explanation:

We need to find the angular velocity in rad/hour when wind is moving with a speed of 4.1 rpm.

We know that,

1 revolution = 2π rad

1 hour = 60 minute

So,

$4.1\ rmm = 4.1\times \dfrac{2\pi }{(\dfrac{1}{60})}\ rad/h\\\\=1545.66\ rad/h$

So, the required angular velocity is 1545.66 rad/h.

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2 years ago

Show that if a and b are positive integers, then ab = gcd (a, b). lcm (a, b) . [Hint: Use prime factorizations of a and b and th

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Answer:

$ab=\gcd(a,b)\cdot \text{lcm}(a,b)$

Step-by-step explanation:

Using the hint, write a and b in the following prime factorization:

$a=p_1^{x_1}p_2^{x_2}\cdots p_t^{x_t}\cdot q,$

$b=p_1^{y_1}p_2^{y_2}\cdots p_t^{y_t}\cdot r,$

where $\gcd(p_i,r)=1,\ \gcd(p_i,q)=1,\ \gcd(r,q)=1,\ \gcd(p_i,p_j)=1,$ for i ≠ j.

Then by the formulae for gcd(a,b) and lcm(a,b) we know that:

$\gcd(a,b)=p_1^{\min(x_1,y_1)}p_2^{\min (x_2,y_2)} \cdots p_t^{\min(x_t,y_t)}$

$\text{lcm}(a,b)= q\cdot r\cdot p_1^{\max(x_1,y_1)}p_2^{\max(x_2,y_2)}\cdots p_t^{\max(x_t,y_t)}$

Note that the expression $\min(x_i,y_i)+\max(x_i,y_i)=x_i+y_i$ for all i, since if the minimum is, <em>without loss of generality</em>, $x_i$ , then the maximum must be $y_i$ , and viceversa. Then, it is straightforward to verify that when we multiply gcd(a, b) and lcm(a, b) its prime factorization matches the prime factorization of ab, and so we can see the equaility holds:

$\gcd(a,b)\cdot \text{lcm}(a,b)=ab.$

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