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bearhunter [10]
3 years ago
15

a man of mass 50 kg climbs up stairs each of height 0.2 m in 20 seconds .calculate the power of the man​

Physics
1 answer:
Delicious77 [7]3 years ago
4 0
How many stairs?
You can use this to find the work
U
W=mgh
And the power by
P=W/T
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Answer:

The correct option is: Total energy

Explanation:

The Hamiltonian operator, in quantum mechanics, is an operator that is associated with the<u> total energy of the system.</u> It is equal to the sum of the total kinetic energy and the potential energy of all the particles of the system.

The Hamiltonian operator was named after the Irish mathematician, William Rowan Hamiltonis denoted and is denoted by H.

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2 years ago
Atoms are the smallest portions of an element that retain the original characteristics of the element. True or false
kozerog [31]

Answer:

True

Explanation:

Going even smaller than atoms would get you to subatomic particles such as quarks. From there, it is impossible to distinguish elements. So, yes, atoms are the smallest portions of an element that retains the original characteristic of the element.

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3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

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3 years ago
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Gelneren [198K]
All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number

Hope this helps!</span>
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3 years ago
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