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Brut [27]
2 years ago
8

5) A girl with 40kg mass is riding an 8 kg bicycle. At the top of the hill her speed is 6.0 m/s, and her speed triples at the bo

ttom of the hill. The hill is 15.0 m high. What is the work done by the non conservative forces?
Physics
1 answer:
Sedaia [141]2 years ago
6 0

Answer:

Work = 6912 joules

Explanation:

Non-conservative forces are dissipative forces such as friction or air resistance. These forces take energy away from the system as the system progresses, energy that you can't get back. These forces are path dependent; therefore it matters where the object starts and stops.

Total mass = 40 + 8 = 48kg

Initial speed u= 6 m/s

Final speed v = 3*initial

Final speed v = 3* 6 = 18 m/s

Distance s = 15

Acceleration a is?

V² = U² + 2aS

18² = 6² + 2a*15

324 = 36 + 30a

324-36= 30a

288 = 30a

288/30 = a

9.6= a

Force = mass* acceleration

Force = 48*9.6

Force = 460.8N

Work = force*distance

Work = 460.8*15

Work = 6912 joules

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If the line spectrum of lithium has a red line at 670.8 nm. How much energy does a a photon with this wavelength have?
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5 0
3 years ago
A 20-lb force acts to the west while an 80-lb force acts 45° east of north. The magnitu 80 lb 70 lb 067 lb 100 lb
olga55 [171]

Answer:

  • The magnitude of the resulting force is 67 lbf.

Explanation:

Taking the east as the positive x direction, and the north as the positive y direction.

The first force points west, this is, in the direction of -\hat{i}, so, is

\vec{F}_1 = - 20 \ lbf \ \hat{j}

\vec{F}_1 = (0 , - 20 \ lbf \)

For the second force, knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector and θ the angle with the positive x direction.

So, the second force is

\vec{F}_2 = 80 \ lbf \ \ ( \ cos(45 \°) \ , \ sin (45 \°) \ )

\vec{F}_2 = ( \ 56.5685 \ lbf \ , \ 56.5685 \ lbf \ )

The net force will be :

\vec{F}_{net} = \vec{F}_1 + \vec{F}_2

\vec{F}_{net} = (0 , - 20 \ lbf \) + ( \ 56.5685 \ lbf \ , \ 56.5685 \ lbf \ )

\vec{F}_{net} =  ( \ 56.5685 \ lbf \ , \ 56.5685 \ lbf \ - 20 \ lbf  )

\vec{F}_{net} =  ( \ 56.5685 \ lbf \ , \ 36.5685 \ lbf \  )

To obtain the magnitude, we can use the Pythagorean Theorem

|\vec{F}_{net}| = \sqrt{F_{net_x}^2 +F_{net_y}^2}

|\vec{F}_{net}| = \sqrt{( \ 56.5685 \ lbf \ )^2 +( \ 36.5685 \ lbf \ )^2}

|\vec{F}_{net}| = 67.36 \ lbf

And this is the magnitude we are looking for.

8 0
3 years ago
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