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Butoxors [25]
3 years ago
15

3. A 64 lb weight stretches a spring 4 ft in equilibrium. The weight is initially displaced 6 inches above equilibrium and given

a downward velocity of 6 ft/sec. Find its displacement for t > 0 if the medium resists the motion with a force equal to 3 times the speed in ft/sec (that is, the damping constant c = 3). Note that there is no external force in this case.

Physics
1 answer:
Damm [24]3 years ago
6 0

Answer:

Explanation:

the solution is given in the attached pictures

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Acceleration is defined as the rate of change for which of the following
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Velocity


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The diagram below shows a boat moving north in a river at 3m/s while the current in the river moves south at 1 m/s.
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Answer it will move slower:

Explanation: cause the force that’s opposite of it

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1pt If the room is the frame of reference for the study of an object moving in three dimensions, from where should distances be
oksian1 [2.3K]

Answer:

Option B, two walls and the floor

Explanation:

The distance should be measured from the point where at least the three axed meet.

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Horizontal wall2 = X Axis

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Measurement of structure in the cosmic microwave background radiation has recently indicated that we live in a flat universe bet
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3 years ago
Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on
Romashka [77]

Answer:

q_1=\pm0.03 \mu C and q_2=\pm0.02 \mu C.

Explanation:

According to Coulomb's law, the magnitude of  force between two point object having change q_1 and q_2 and by a dicstanced is

F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)

Where, \epsilon_0 is the permitivity of free space and

\frac{1}{4\pi\spsilon_0}=9\times10^9 in SI unit.

Before  dcollision:

Charges on both the sphere are q_1 and q_2, d=20cm=0.2m, and F_c=1.35\times10^{-4} N

So, from equation (i)

1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}

\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

\frac{q_1+q_2}{2}

d=20cm=0.2m, and F_c=1.406\times10^{-4} N

So, from equation (i)

1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}

\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}

\Rightarrow q_1+q_2=\pm5\times 10^{-8}

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)

\Rightarrow q_1=5\times 10^{-8}-q_2

The equation (ii) become:

(5\times 10^{-8}-q_2)q_2=6\times10^{-16}

\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0

\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}

From equation (iii)

q_1=2\times10^{-8}, 3\times10^{-8}

So, the magnitude of initial charges on both the sphere are 3\times10^{-8} Coulombs=0.03 \mu C and 2\times10^{-8} Colombs or 0.02 \mu C.

Considerion the nature of charges too,

q_1=\pm0.03 \mu C and q_2=\pm0.02 \mu C.

4 0
3 years ago
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