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2 years ago

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Write an essay about Paralympic and Special Olympic Essay 100 points

1 answer:

wlad13 [49]2 years ago

8 0

Answer:

The Paralympics does the same. The Special Olympics and the Paralympics are both sports organizations for disabled athletes, but they are very different organizations and are not connected. Many people confuse the Paralympics and the Special Olympics.

Explanation:

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PLZ help this should be super easy

Ugo [173]

The answer is number (3) if am wrong am sry

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2 years ago

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Between 1911 and 1990, the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of 1.2 mm/y.T

Anit [1.1K]

Answer:

Angular speed, $\omega=6.90\times 10^{-13}\ rad/s$

Explanation:

It is given that,

The top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of, v = 1.2 mm/yr

$1\ mm/yr=3.171\times 10^{-11}\ m/s$

Velocity, $v=3.80\times 10^{-11}\ m/s$

Height of the tower, h = 55 m

The height of the tower is equivalent to the radius. Let $\omega$ is the angular speed of the tower’s top about its base. The relation between the angular speed and the angular speed is given by :

$v=r\omega$

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{3.80\times 10^{-11}\ m/s}{55\ m}$

$\omega=6.90\times 10^{-13}\ rad/s$

So, the average angular speed of the tower’s top about its base is $6.90\times 10^{-13}\ rad/s$ . Hence, this is the required solution.

6 0

3 years ago

A 150-kg object takes 1.5 minutes to travel a 2,500-meter straight path. It begins the trip traveling 120 m/s and decelerates to

podryga [215]

I believe it is -1.11 m/s^2. I will let you know if its correct

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3 years ago

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A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

2 years ago

PLEASE HELP!!! WOULD REALLY APPRECIATE!

Eva8 [605]

Answer:

a. slope=rise/run

rise=0.02

run=-2

determined using the point (3,0.08) and (1,0.1) on the graph

slope=0.02/-2

= -0.01 or -1/100

b.area= area of trapizoid+ rectangle

((0.07+0.11)÷2)×4+1×0.07

0.36+0.07

=0.43$

c. the area represent the total cost after 5 hours

PLEASE MARK BRAINLIEST

3 0

1 year ago