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3 years ago

How much would a 15.0 kg object weigh on Neptune?

Physics

2 answers:

yKpoI14uk [10]3 years ago

6 0

Weight = (mass) · (gravity)

Weight = (15.0 kg) · (11.2 m/s²)

Weight = (15.0 · 11.2) · (kg · m/s²)

Weight = 168 Newtons

yan [13]3 years ago

5 0

The gravity on Neptune is 11.15 m/s²
the gravity on earth is 9.81 m/s²
divide the Neptune and earth gravity we get 1.13
which means object on neptune is 1.13 heavier than earth
yield, weigh of the object on neptune is 1.13×15=17.04kg

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Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When

kupik [55]

Answer:

therefore critical angle c= 69.79°

Explanation:

Canola oil is less dense than water, so it floats over water.

Given $n_{canola}= 1.47$

which is higher than that of water

refractive index of water $n_{water}=1.33$

to calculate critical angle of light going from the oil into water

we know that

$sinc= \frac{n_{water}}{n_{canola}}$

now putting values we get

$sinc= \frac{1.33}{1.47}$

c= $sin^{-1}(\frac{1.33}{1.47} )$

c=69.79°

therefore critical angle c= 69.79°

8 0

3 years ago

Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i

spayn [35]

Answer:

$v = 15.56 m/s$

$v = 56 km/h$

Explanation:

When coefficient of friction is approximately zero then we have

$F_ncos\theta = mg$

$F_n sin\theta = \frac{mv^2}{R}$

$tan\theta = \frac{v^2}{Rg}$

here we know that

$v = 40 km/h = 11.11 m/s$

R = 30 m

$tan\theta = \frac{11.11^2}{30\times 9.81}$

$\theta = 22.75 degree$

now when friction coefficient is 0.30 then we have

$F_n cos\theta = mg + F_f sin\theta$

$F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}$

now we have

$v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}$

$v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}$

$v = 15.56 m/s$

$v = 56 km/h$

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Katarina [22]

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3 years ago