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spin [16.1K]
3 years ago
6

A diver running 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. how hi

gh was the cliff and how far from its base did the diver hit the water?
Physics
1 answer:
Elodia [21]3 years ago
5 0

<u>Answer:</u>

a) Height of cliff = 44.145 meter

b) 7.5 meter far from its base the diver hit the water.

<u>Explanation:</u>

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Diver's vertical motion:

   Initial velocity = 0 m/s, acceleration = 9.8 m/s^2, we need to calculate displacement when time is 3 seconds.

  s= ut+\frac{1}{2} at^2\\ \\ =0*3+\frac{1}{2} *9.81*3^2\\ \\ =44.145meter

 So height of cliff = 44.145 meter.

b) Diver's horizontal motion:

Initial velocity = 2.5 m/s, acceleration = 0 m/s^2, we need to calculate displacement when time is 3 seconds.

s= 2.5*3+\frac{1}{2}*0*3^2\\ \\ =7.5meter

So 7.5 meter far from its base the diver hit the water.

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For other statements:

Batteries donot store electric charge but they store chemical energy

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What is special about nitrogen, and what is its main function in the atmosphere
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Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

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2 years ago
Can molecules with double or triple bonds twist
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No.

Explanation:

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6 0
3 years ago
A machine runs for 50 seconds with a steady power output of 100 watts. How many joules of work does the
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Answer:

The answer to your question is when time = 50 s, work = 5000 J

                                                    when time = 90 s, work = 9000 J

Explanation:

Data

time = 50 s or 90 s

Power = 100 watts

Power is defined as the rate of work done per unit of time.

           Power = Work / time

-Solve for Work

           Work = Power x time

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-Result

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2.-When time = 90 s

           Work = 100 x 90

-Result

          Work = 9000 watts

6 0
2 years ago
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