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photoshop1234 [79]
3 years ago
8

A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee

n them if their centers are 0.5 m apart?
1.0 × 10–13 m
1.3 × 10–10 m
2.6 × 10–10 m
1.1 × 10–8 m
Physics
2 answers:
larisa [96]3 years ago
8 0
We can use the equation for Newton's Law of Gravitation
Fg = (Gm₁m₂)/r²
Where gravitational constant = G = 6.674 x 10⁻¹¹ N · m²/kg²
mass m₁ = 0.145 kg
mass m₂ = 6.8 kg
distance between centers of masses = r = 0.5 m

Substitute these values into... 
Fg = (Gm₁m₂)/r²
Fg = ((6.674 x 10⁻¹¹)(0.145)(6.8)) / (0.5)²
Fg = 2.63 x 10⁻¹⁰ N 

Therefore, your answer should be 
<span>2.6 × 10–10</span>
Brums [2.3K]3 years ago
8 0

Answer:

C: 2.6 × 10^–10 N

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Answer:

\theta = 76.9 degree

Explanation:

As we know that the resultant of two vectors is given as

R = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 cos\theta

here we know that

R = 87 Lb

F_1 = 41 Lb

F_2 = 68 Lb

now we have

87 = \sqrt{41^2 + 68^2 + 2(41)(68)cos\theta

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\theta = cos^{-1}(\frac{1264}{5576})

\theta = 76.9 degree

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The frequency of the second harmonic of a certain musical instrument is 100 Hz. What is the fundamental frequency of the instrum
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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
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