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3 years ago

One of the Lady Spartans was falling to the ground after

Physics

1 answer:

dem82 [27]3 years ago

0 0

Answer:

Explanation:

Given that,

A lady falling has a final velocity of 4m/s

v = 4m/s

Mass of the lady is 60kg.

m = 60kg

Using conservation of energy, the potential energy of the body from the point where the lady is dropping is converted to the final kinetic energy of the lady.

Therefore,

P.E = K.E(final) = ½mv²

P.E = ½ × 60 × 4²

P.E = 480 J.

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If light of wavelength 700 nm strikes such a photocathode, what will be the maximum kinetic energy, in eV , of the emitted elect

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If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.

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threshold when KE = 0

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What is photocathode?

A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
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What topic in a biology textbook is directly related to physics

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4 years ago

A man is sitting on a chair with wheels. He grabs a 2.1 kg book from the desk and throws

geniusboy [140]

The speed of the man and the chair after the book is thrown is 0.2 m/s.

The given parameters:

mass of the book, m₁ = 2.1 kg
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The total mass of the man and the book is calculated as follows;

m₂ = 70 kg + 9.2 kg

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$m_1 u_1 = m_2u_2\\\\u_2 = \frac{m_1u_1}{m_2} \\\\u_2 = \frac{2.1 \times 7.2}{79.2} \\\\u_2 = 0.2 \ m/s$

Thus, the speed of the man and the chair after the book is thrown is 0.2 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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3 years ago

Determine the force per meter on a lightning bolt at the equator that carries 20 000 A and is perpendicular to the Earth’s magne

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To determine the force per meter of a light bolt at the equator that carries 20 000 A and is perpendicular to the Earth's magnetic field with a magnitude of 3x10-5 T, we do this:
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3 years ago

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A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$

Calculate the time taken by the stone to reach the ground.

$t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s$

The horizontal distance traveled by the stone is given by,

$R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m$

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

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3 years ago