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Dennis_Churaev [7]
3 years ago
7

Determine the kinetic energy of a 55kg woman running with the velocity of 5.87m/s

Physics
1 answer:
Greeley [361]3 years ago
4 0
The formula for kinetic energy = ½m·v<span>2

1/2 * 55 kg x 5,87 m/s ^2 = 27.5 x </span>34.4569 = <span>947.56475 Joule </span>≈ 948 J
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Which of these is true about kinetic energy but not necessarily true about potential energy
cestrela7 [59]

Kinetic energy is never negative, but potential energy can be.

Potential energy depends on height above some reference level,
and you can pick any level you want as the reference.  So, if the
object is below the reference level you pick, then its potential
energy relative to your reference level is negative. 

What that means is:  You have to lift it / do work on it / give it more
energy than it has now ... in order to move it to the reference level.

(That's exactly the situation with electrons bound to an atom.  Their
energy is considered negative, because we have to do work and
give them more energy to rip them away from the atom.)
_____________________________________

Regarding the other choices:

-- Kinetic energy is scalar ... Yes.  So is potential energy.

-- Kinetic energy increases with height ...
   No. It doesn't, but potential energy does.

-- Kinetic energy depends on position ...
   No. It doesn't, but potential energy does.

3 0
3 years ago
Smoking and your health help!!!!
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4 0
3 years ago
The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

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3 0
3 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

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Oduvanchick [21]

Answer:

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