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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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3 years ago

A snail con move at 3cm/s if it’s ke is 300joules what it it’s mass?

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3 years ago

What is the magnetic potential energy stored in a cylindrical volume of height hcylin = 50 mm and radius Rcylin = 24 mm that sym

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Answer:

$U=1.02\times 10^{-7}\ J$

Explanation:

Given that

h= 50 mm = 0.05 m

R= 24 mm = 0.024 m

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I= 2.9 A

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$B=\dfrac{\mu _oI}{2\pi r}$

Volume of small element

dV= 2πr hdr

Now the magnetic potential energy given as

$U=\int \dfrac{B^2}{2\mu_o}dV$

$B=\dfrac{\mu _oI}{2\pi r}$

dV= 2πr hdr

$U=\int\dfrac{\left(\dfrac{\mu _oI}{2\pi r}\right)^2}{2\mu_o}2\pi rhdr$

$U=\dfrac{\mu _oI^2h}{4\pi}\int_{R_w}^{R}\dfrac{dr}{r}$

$U=\dfrac{\mu _oI^2h}{4\pi}\times \ln\dfrac{R}{R_w}$

Now by putting the values

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3 years ago

The angular speed of an automobile engine is increased at a constant rate from 1260 rev/min to 3460 rev/min in 9.90 s. (a) What

larisa86 [58]

Answer:

13333.33 rev/min²

Explanation:

Given:

Initial angular speed of the automobile (ω₁) = 1260 rev/min

Final angular speed of the automobile (ω₂) = 3460 rev/min

Time interval for the change in speed (t) = 9.90 s

Angular acceleration of the automobile (α) = ?

Consider the sense of rotation to be positive. So, making use of the equation of motion of rotation, we have:

$\omega_2=\omega_1+\alpha t$

Rewriting in terms of 'α', we get:

$\alpha =\dfrac{\omega_2-\omega_1}{t}$

Converting time 't' from seconds to minutes using conversion factor.

1 sec = $\frac{1}{60}\ min$

So, 9.90 s = $9.9\times \frac{1}{60}=0.165\ min$

Plug in all the given values and solve for 'α'. This gives,

$\alpha =\frac{3460-1260}{0.165}\ rev/min^2\\\\\alpha=\frac{2200}{0.165}=13333.33\ rev/min^2$

Therefore, the angular acceleration in revolutions per minute squared is 13333.33 rev/min².

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3 years ago