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Nutka1998 [239]

3 years ago

12

Kgms^-1 ÷ m× (ms^-1)^2 Can u simplify this

1 answer:

Ymorist [56]3 years ago

7 0

Answer:

(kgm^2) / (s^3)

Step-by-step explanation:

Comment if you want steps.

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The saucer for Aisha's teacup has a diameter of 6 inches. What is the saucer's radius?

Aleonysh [2.5K]

I think its 6 but im not sure to be exactly right

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2 years ago

To which sets of numbers - 1/4 does<br> belong?

Afina-wow [57]

Where are the sets? I don't see any...

6 0

2 years ago

Solve: -6 (-2)÷4(-3)

Katen [24]

Answer:

-9

Step-by-step explanation:

Simplify the following:

-6×(-2)/4 (-3)

-6×(-2)/4 (-3) = (-6 (-2) (-3))/4:

(-6 (-2) (-3))/4

The gcd of -2 and 4 is 2, so (-6 (-2) (-3))/4 = (-6 (2 (-1)) (-3))/(2×2) = 2/2×(-6 (-1) (-3))/2 = (-6 (-1) (-3))/2:

(-6-1 (-3))/2

(-6)/2 = (2 (-3))/2 = -3:

--3 (-3)

-3 (-1) = 3:

3 (-3)

3 (-3) = -9:

Answer: -9

6 0

3 years ago

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago

Match each expression to its equivalent standard form

pogonyaev

Move the first one to the bottom,then the second to the top ans you will know where to put the rest

8 0

3 years ago

Read 2 more answers