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3 years ago

9

A ball is thrown directly upward into the air. The graph below shows the vertical position of the ball with respect to time

1 answer:

Alexxx [7]3 years ago

5 0

.Explanation:

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Which value is equivalent to 178 centimeters?

IRINA_888 [86]

178 centimeters=1.78 meters
which means that the answer is C.

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3 years ago

Which correctly lists three main causes of deforestation?

ASHA 777 [7]

I think it’s A. Mining and logging are huge factors

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3 years ago

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the tallest man made structure At present is theWarszawa radio massin Warsaw Poland.This radio mast rises 646m above ground near

PtichkaEL [24]

All of that fluff at the beginning is interesting, but completely irrelevant
to the question. The question is just asking for the mass of an object
that weighs 3.6N on Earth.

Weight = (mass) x (acceleration of gravity)

3.6N = (mass) x (9.8 m/s²)

Divide each side
by 9.8 m/s : Mass = 3.6N / 9.8 m/s² = <em>0.367 kilogram</em> (rounded)

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3 years ago

Terry can ride 30 miles in 2 hours. If his riding speed is

marishachu [46]

Answer:

Explanation:

Terry can ride at a speed of

V = 30 miles in 2hours

Speed = distance / time

V = 30 /2

V = 15 mile/hour

So, we want to know the distance traveled in 1.7hours

Then,

Speed = distane / time

Distance = speed × time

Distance =15 × 1.7

Distance = 25.5 miles

So, the distance traveled in 1.7hours is 25.5 miles

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3 years ago

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A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of

LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

$\Psi=4RcE\pi$

Where

$\Psi =$ The energy fluency

c = Activity of the source

r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

$I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}$

Then replacing our values we have that

$I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}$

$I = 1.3*10^7 Bq$

We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.

Now the energy fluency would be,

$\Psi = \frac{cE}{4\pir^2}$

$\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}$

$\Psi = 2.14*10^4 MeV/cm^2.s$

The uncollided flux density at the outer surface of the tank nearest the source is $\Psi = 2.14*10^4 MeV/cm^2.s$

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3 years ago