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Ad libitum [116K]

3 years ago

13

The asteroid Ceres has a mass 6.797 × 1020 kg and a radius of 472.9 km. What is g on the surface? The value of the universal gra

vitational constant is 6.67259 × 10−11 N · m2 /kg2 . Answer in units of m/s 2 .

1 answer:

raketka [301]3 years ago

8 0

Answer:

$g=0.20\ m/s^2$

Explanation:

It is given that,

Mass of the asteroid Ceres, $m=6.797\times 10^{20}\ kg$

Radius of the asteroid, $r=472.9\ km=472.9\times 10^3\ m$

The value of universal gravitational constant, $G=6.67259\times 10^{-11}\ N.m^2/kg^2$

We know that the expression for the acceleration due to gravity is given by :

$g=\dfrac{Gm}{r^2}$

$g=\dfrac{6.67259\times 10^{-11}\times 6.797\times 10^{20}}{(472.9\times 10^3)^2}$

$g=0.20\ m/s^2$

So, the value of acceleration due to gravity on that planet is $0.20\ m/s^2$ . Hence, this is the required solution.

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3 years ago

A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

Blababa [14]

Answer:

(B) 1.6 m/s^2

Explanation:

The equation of the forces acting on the box in the direction parallel to the slope is:

$mg sin \theta - \mu N = ma$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, $\theta=39^{\circ}$ being the angle of the incline

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a is the acceleration

The equation of the force along the direction perpendicular to the slope is

$N-mg cos \theta =0$

where $mg cos \theta$ is the component of the weight in the direction perpendicular to the slope. Solving for N,

$N=mg cos \theta$

Substituting into (1), solving for a, we find the acceleration:

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6 0

4 years ago

Please need some help on this thank you so much

fomenos

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4 years ago

Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a

lilavasa [31]

Answer:

$\lambda = 570\ nm$

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

$tan \theta = \dfrac{y}{L}$

$tan \theta = \dfrac{4}{4\times 10^3}$

$\theta = 0.0572$

Now.

$W sin \theta = m \lambda$

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$5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda$

$\lambda = 569.99 \times 10^{-9}\ m$

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4 years ago

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nataly862011 [7]

Answer:

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4 years ago