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3 years ago

Two spherical objects have masses of 100 kg and 200 kg. Their centers are

Physics

1 answer:

drek231 [11]3 years ago

6 0

Answer:

8.34 x 10⁻⁶N

Explanation:

Given parameters:

Mass 1 = 100kg

Mass 2 = 200kg

Distance of separation = 40cm = 0.4m

Unknown:

Gravitational force of attraction between them = ?

Solution:

To solve this problem, we use the expression below which is derived from the Newton's law of universal gravitation:

Fg = $\frac{G x mass 1 x mass 2}{d^{2} }$

G is the universal gravitation constant = 6.67 x 10⁻¹¹

d is the separation

Now;

Fg = $\frac{6.67 x 10^{-11} x 100 x 200}{0.4^{2} }$ = 8.34 x 10⁻⁶N

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How many Sig Figs (Significant Figures) are in each number? 5070.0 870.064080

nlexa [21]

∑ Hey, Lethality ⊃

Answer:

5070.0 has 5 significant figures

870.064080 has 9 significant figures

Explanation:

Given:

How many Sig Figs (Significant Figures) are in each number?

5070.0

870.064080

Solution:

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5070.0

5070.0 has 5 significant figures ( 5 , 0 , 7 , 0 , and 0 )

Number of significant figures: 5

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870.064080

870.064080 has 9 significant figures ( 8, 7 ,0,0, 6,4,0,8 and 0 )

Number of significant figures: 9

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xcookiex12

8/23/2022

5 0

1 year ago

Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea

Alexxx [7]

Answer:

acceleration are

hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

Em₀ = U = m g h

Final point. To get off the ramp

Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

Em₀ = Em_f

mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

v = w r

w = v / r

we substitute

mg h = ½ v² (m + I / r²)

v² = 2 gh $\frac{m}{m+ \frac{I}{r^2} }$

v² = 2gh $\frac{1}{1 + \frac{I}{m r^2} }$

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

v² = v₀² + 2 a L

where L is the length of the plane

v² = 2 a L

a = v² / 2L

we substitute

a = $g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }$

let's use trigonometry

sin θ = h / L

we substitute

a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

I = m r²

we look for the acerleracion

a₁ = g sin θ $\frac{1}{1 + \frac{mr^2 }{m r^2 } }$ 1/1 + mr² / mr² =

a₁ = g sin θ ½

b) solid cylinder

I = ½ m r²

a₂ = g sin θ $\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }$ = g sin θ $\frac{1}{1+ \frac{1}{2} }$

a₂ = g sin θ ⅔

c) hollow sphere

I = 2/3 m r²

a₃ = g sin θ $\frac{1}{1 + \frac{2}{3} }$

a₃ = g sin θ $\frac{3}{5}$

d) solid sphere

I = 2/5 m r²

a₄ = g sin θ $\frac{1 }{1 + \frac{2}{5} }$

a₄ = g sin θ $\frac{5}{7}$

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ $\frac{105}{210}$