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3 years ago

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The acceleration due to gravity at sea level is g=9.81 m/s^2. The radius of the earth is 6370 km. The universal gravitational co

nstant is G = 6.67 × 10−11 N-m^2/kg^2. Use this information to determine the mass of the earth.

1 answer:

solmaris [256]3 years ago

3 0

Answer:

Mass of earth will be $M=5.96\times 10^{24}kg$

Explanation:

We have given acceleration due to gravity $g=9.81m/sec^2$

Radius of earth = 6370 km = $6370\times 10^3m$

Gravitational constant $G=6.67\times 10^{-11}Nm^2/kg^2$

We know that acceleration due to gravity is given by

$g=\frac{GM}{R^2}$ , here G is gravitational constant, M is mass of earth and R is radius of earth

So $9.81=\frac{6.67\times 10^{-11}\times M}{(6370\times 10^3)^2}$

$M=5.96\times 10^{24}kg$

So mass of earth will be $M=5.96\times 10^{24}kg$

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Answer:

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For block B, a = 15 ft/s²

Explanation:

A free body diagram for this force system is attached to this solution

Mass of block A = m₁ = 8 lb

Mass of block B = m₂ = 6 lb

Coefficient of kinetic friction = μ

Normal reaction on the blocks = N

Spring stiffness of the spring btw block A and B = k = 20 lb/ft

Compression of the spring = 0.2 ft

Analysing Block A first

The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

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Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 8/32.2 = 0.248 lbm

(20×0.2) - (0.2 × 8) = (0.248) aₓ

aₓ = 9.66 ft/s²

The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

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(k × x) - (μ × N) = maₓ

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aₓ = 15 ft/s²

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Answer:

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Where $v_{o}$ is the initial velocity of the particle, measured in meters per second.

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$s(t) = \int {v(t)} \, dt$

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$s_{o} = -5$

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