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prisoha [69]
4 years ago
8

You and your friends find a rope that hangs down 16m from a high tree branch right at the edge of a river. You find that you can

run, grab the rope, swing out over the river, and drop into the water. You run at 2.0 m/s and grab the rope, launching yourself out over the water. How long must you hang on if you want to drop into the water at the greatest possible distance from the edge?
Physics
1 answer:
Harlamova29_29 [7]4 years ago
4 0

Answer:

t = 2\ s

Explanation:

given,

length of the rope = 16 m

speed of the man = 2 m/s

using the formula of time period

T =2 \pi \sqrt{\dfrac{L}{g}}

T =2 \pi \sqrt{\dfrac{16}{9.8}}

T = 8.028\ s

To cover the maximum distance you need to leave the when the rope is shows maximum displacement.

To reach the displacement time to leave the rope is one fourth of the time period.

t = \dfrac{T}{4}

t = \dfrac{8.03}{4}

t = 2\ s

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a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction  30° east of north
liq [111]

When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

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So, we first find the components of the red arrow along X and Y.

Component of D_{2} along X-axis is given by  D_{2x}  = D_{2} Sin 30 = 10 km

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We now add all the vectors along X and along Y separately.

Net Displacement along X  D_{netX} = 10 km

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Now that we have the components of the net displacement along X and Y, we make use of Pythagorean Theorem to calculate the D_{net}

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6 0
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A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it
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Two charged spheres on a frictionless horizontal surface are attached to opposite ends of a string &amp; are in static equilibri
deff fn [24]

Answer:

the red sphere has a charge of 9.899 10⁻⁶ C and

the green sphere 0.101 10⁻⁶ C

Explanation:

For this exercise we will use coulomb's law

          F = k q₁ q₂ / r²

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they also indicate the value of the total load

         q₁ + q₂ = 10 10⁻⁶ C

we substitute the values

         2.5 = 9 10⁹ q₁ q₂ / 0.06²

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we have two equations with two i unknowns, so the system can be solved.

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we substitute in the other equation

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let's solve the equation

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          q₂ = 0.9899 10⁻⁵ C

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let's find the charge of the other sphere

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therefore the red sphere has a charge of 9.899 10⁻⁶ C and

the green sphere 0.101 10⁻⁶ C

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