State whether 0 is a rational number

A 748-N man stands in the middle of a frozen pond of radius 4.0 m. He is unable to get to the other side because of a lack of fr

Ber [7]

Answer:

The man will take 64 seconds to reach to the south shore of the frozen pond.

Step-by-step explanation:

Given:

Weight of the man = 748 N Mass of the man, $(m)= \frac{W}{g}$ = $\frac{748}{9.8}$ = $76.32$ kg

Radius of the pond $(r)$ = 4 m

Mass of the textbook = 1.2 kg

Velocity at which the textbook is thrown = 4 ms^1

We have to find the velocity of the man after the throw.

Let the velocity is $V_m$ .

Now using law of conservation of momentum we can find the $V_m$ value.

$m_(_b_)V_b_(_i_) +m_(_m_)V_m_(_i_) =m_(_b_)V_b_(_f_)+m_(_m_) V_m_(_f_)$

Considering $V_m_(_f_)=V_m$

And initial velocity of both the man and book i.e $V_b_(_i_)=0,\ V_m_(i_)=0$

So,

⇒ $0 =m_(_b_)V_b_(_f_)+m_(_m_) V_m$

⇒ Plugging the values.

⇒ $V_m=-\frac{m_(_b_)V_b_(_f_)}{m_(_m_)}$

⇒ $V_m=-\frac{1.2\times 4}{76.32}$

⇒ $V_m=-0.062$ ms^-1

Here the negative velocity is meant for opposite direction of the throw.

Numerically we will write, $V_m = 0.062$

With this velocity the man will move towards south.

We have to calculate the time taken by the man to move to its south shore.

And we know $velocity(v)\times time(t) = distance(d)$

Let the time taken be $t$ and $v\times t = d$ and $d=r$ then, $V_m\times t=r$

Then

⇒ $t=\frac{radius\ (r)}{V_m}$

⇒ Plugging the values.

⇒ $t=\frac{4}{0.062}$

⇒ $t =64$ sec

The man will take 64 seconds to reach to the south shore of the frozen pond (circular).

5 0

3 years ago

6. Find LCM (10,14,63).

Katen [24]

Answer:

630 is the LCM

Step-by-step explanation:

You have to list all the multiples of each number till you find the lowest same value one which in this case it's 630.

Multiples of 10:

10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 490, 500, 510, 520, 530, 540, 550, 560, 570, 580, 590, 600, 610, 620, 630, 640, 650

Multiples of 14:

14, 28, 42, 56, 70, 84, 98, 112, 126, 140, 154, 168, 182, 196, 210, 224, 238, 252, 266, 280, 294, 308, 322, 336, 350, 364, 378, 392, 406, 420, 434, 448, 462, 476, 490, 504, 518, 532, 546, 560, 574, 588, 602, 616, 630, 644, 658

Multiples of 63:

63, 126, 189, 252, 315, 378, 441, 504, 567, 630, 693, 756

3 0

3 years ago

What is 4n + 3 ( n – 5 ) + 2 simplified? I suck At math sry

Tema [17]

Answer:

$7n−13$

Step-by-step explanation:

$4n+3(n−5)+2$

$4n+3n+3 \times (−5+2)$

$4n+3n−15+2$

$7n−15+27n-15+2$

$7n−13$

<h3>Hope it is helpful....</h3>

4 0

3 years ago

Read 2 more answers

X f(x) 2 18 3 27 4 36 5 45

Ad libitum [116K]

Answer:

$f(x) = 9x$

Step-by-step explanation:

18=9×2

27=3*9

36=4*9

45=5*9

5 0

2 years ago

The length of a photograph is 1 cm less than twice the width. The area is 91cm^2. find the width and length.

maksim [4K]

Answer:

Width=6.5 cm

Length=12 cm

Step-by-step explanation:

Step 1: Express the lengths and widths

Width=w

Length=l, but 1 cm less than twice the width=(2×w)-1=2 w-1

Step 2: Solve for the length and width

A=L×W

where;

A=area of the photograph

L=length of the photograph

W=width of the photograph

In our case;

A=91 cm²

L=2 w-1

W=w

91=(2 w-1)w

2 w²-w=91

2 w²-w-91=0, is a quadratic equation

solve for w

w={-1±√(-1²-4×2×-91)}/(2×2)

w=(-1±27)/4

w=(27-1)/4=6.5, or (-1-27)/4=-8

Take w=6.5 cm

L=(2×6.5)-1=13-1=12 cm

Width=6.5 cm

Length=12 cm

6 0

3 years ago

State whether 0 is a rational number​

State whether 0 is a rational number