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mars1129 [50]

Answer:

16

Step-by-step explanation:

separate into 4 triangles. 2x1= 2

triangle area formula= 1/2 l x w

3x4=12 12+4 is 16

7 0

3 years ago

A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a

AnnZ [28]

<span>Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed:
a = omega^2*r

omega and r in terms of given data:
omega = 2*Pi/T
r = d/2

Thus:
a = 2*Pi^2*d/T^2

What forces cause this acceleration for the passenger, at either top or bottom?

At top (acceleration is downward):
Weight (m*g): downward
Normal force (Ntop): upward

Thus Newton's 2nd law reads:
m*g - Ntop = m*a

At top (acceleration is upward):
Weight (m*g): downward
Normal force (Nbottom): upward

Thus Newton's 2nd law reads:
Nbottom - m*g = m*a

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame:
Ntop = m*(g - a)
Nbottom = m*(g + a)

Substitute a:
Ntop = m*(g - 2*Pi^2*d/T^2)
Nbottom = m*(g + 2*Pi^2*d/T^2)

We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground):
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g)
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g)

Simplify:
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2)
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2)

Data:
d:=22 m; T:=12.5 sec; g:=9.8 N/kg;

Results:
Ntop/(m*g) = 71.64%...she feels "light"
Nbottom/(m*g) = 128.4%...she feels "heavy"</span>

7 0

4 years ago

Read 2 more answers

Hi, teacher I was absent these days and I didn’t understand anything about this lesson and I need help this is not count as a te

motikmotik

Given:

There are given that the cos function:

$cos210^{\circ}=-\frac{\sqrt{3}}{2}$

Explanation:

To find the value, first, we need to use the half-angle formula:

So,

From the half-angle formula:

$cos(\frac{\theta}{2})=\pm\sqrt{\frac{1+cos\theta}{2}}$

Then,

Since 105 degrees is the 2nd quadrant so cosine is negative

Then,

By the formula:

$\begin{gathered} cos(105^{\circ})=cos(\frac{210^{\circ}}{2}) \\ =-\sqrt{\frac{1+cos(210)}{2}} \end{gathered}$

Then,

Put the value of cos210 degrees into the above function:

So,

$\begin{gathered} cos(105^{\circ})=-\sqrt{\frac{1+cos(210)}{2}} \\ cos(105^{\operatorname{\circ}})=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ cos(105^{\circ})=-\sqrt{\frac{2-\sqrt{3}}{4}} \\ cos(105^{\circ})=-\frac{\sqrt{2-\sqrt{3}}}{2} \end{gathered}$

Final answer:

Hence, the value of the cos(105) is shown below:

$cos(105^{\operatorname{\circ}})=-\frac{\sqrt{2-\sqrt{3}}}{2}$

4 0

1 year ago

What is a solutions of x^2-2x+10=0?

Debora [2.8K]

Answer:

x = 1± 3i

Step-by-step explanation:

x^2-2x+10=0

We can complete the square to solve by subtracting 10 from each side

x^2-2x+10-10=-10

x^2 -2x = -10

We need to add (2/2) ^2 to each side or 1

x^2 -2x+1 = -10 +1

x^2 -2x+1 = -9

The left side factors into (x- (2/2) ) ^2

(x-1) ^2 = -9

Take the square root of each side

sqrt((x-1) ^2 =± sqrt(-9)

x-1 = ±sqrt(-1) sqrt(3)

Remember the sqrt(-1) = i

x-1 = ± 3i

Add 1 to each side

x-1+1 = 1± 3i

x = 1± 3i

8 0

3 years ago

What is 1/3 of 1/9 of a ton?<br><br> Make sure your answer is fully reduced

natulia [17]

Answer:

0.03703703703

Step-by-step explanation:

6 0

3 years ago