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igor_vitrenko [27]
3 years ago
12

The current in a river is 1.0 meters/second. Britney swims 300 meters against the current. If she normally swims with a speed of

2.0 meters/second in still water, how long does it take her to complete the trip?
Physics
2 answers:
kotegsom [21]3 years ago
8 0

Answer:

t = 300 seconds

Explanation:

It is given that,

Speed of the current in the river, v_1=1\ m/s

Distance covered by Britney, d = 300 m

Speed of Britney, v_2=2\ m/s

It is mentioned that Britney swims against the current. Her downstream speed is given by :

v=v_2-v_1

v=2-1

v = 1 m/s

Let t is the time taken by her to complete her trip. t is given by :

t=\dfrac{d}{v}

t=\dfrac{300\ m}{1\ m/s}

t = 300 seconds

So, the time taken by her the complete her trip is 300 seconds. Hence, this is the required solution.

Nikitich [7]3 years ago
4 0

The answer is 300s I believe.

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At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 40.0 N at the s
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Answer

acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²

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acceleration due to gravity on earth, g = 9.8 m/s²

a) Mass on the earth surface

    M = \dfrac{W}{g}

    M = \dfrac{40}{9.8}

           M = 4.08 Kg

b) Mass on the surface of Lo

 Mass of an object remain same.

  Hence, mass of object at the surface of Lo = 4.08 Kg.

c) Weight at the surface of Lo

   W' = m g'

   W' =4.08 x 1.81

   W' = 7.38 N

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2 years ago
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7 0
3 years ago
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0
3 years ago
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