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3 years ago

8

Steve want to catch up to Cami, a girl he likes. If he is jogging at 8 m/s and she is walking at 1 m/s. How long will it take hi

m to reach her if they are 10 meters apart?

1 answer:

Aleks04 [339]3 years ago

5 0

Answer:

<u><em>1) if they are moving away from each other it will take 1.43 secs</em></u>

<u><em>2) if they are moving towards each other then it will take 1.11 secs</em></u>

Explanation:

Distance between them is 10 m

Speed ( if they are moving towards each other)= distance/time

time = 10/8+1

time = distance / speed= 10/9= 1.11 secs

if they are moving away from each other than it will take

time = 10/8-1= 10/7= 1.43 secs

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HOW MANY PIES DOSE IT TAKE TO GET TO THE MOON

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i do not have an answer because it depends on the size and the distance lol

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3 years ago

What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?

ivann1987 [24]

Answer:

$2.46\cdot 10^5 J$

Explanation:

The energy of a single photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength

For the photon in this problem,

$\lambda=486 nm=4.86\cdot 10^{-7}m$

So, its energy is

$E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J$

One mole of photons contains a number of photons equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

So, the total energy of one mole of photons is

$E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J$

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3 years ago

What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300

Vinvika [58]

The efficiency of an ideal Carnot heat engine can be written as:
$\eta = 1- \frac{T_{cold}}{T_{hot}}$
where
$T_{cold}$ is the temperature of the cold region
$T_{hot}$ is the temperature of the hot region

For the engine in our problem, we have $T_{cold}=300 K$ and $T_{hot}=400 K$ , so the efficiency is
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3 years ago

A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between

elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

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3 years ago

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart

hichkok12 [17]

Answer:

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Explanation:

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so we will have

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now we have

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so we will have

$r = 3.46 \times 10^8 m$

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$-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5$

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3 years ago