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maksim [4K]
2 years ago
3

A 2-kg stone falls 100 meters near the surface of the earth. it strikes the ground without any rebound thereby making a complete

ly inelastic collision with the earth. approximately how much kinetic energy is transferred to the earth in this process?
Physics
2 answers:
VikaD [51]2 years ago
6 0

Answer:

E = 1960 J

Explanation:

When a stone of mass 2 kg falls from initial height H = 100 m then its total potential energy will convert into kinetic energy just before it will hit the ground

So we will say

mgH = \frac{1}{2}mv^2

so we will have

v = \sqrt{2gH}

now we know that when collision is completely inelastic collision then the mechanical energy is completely lost in the form of thermal energy and other forms of energy

so here total kinetic energy is transferred to the Earth

so we have

E = mgH

E = 2(9.8)100

E = 1960 J

never [62]2 years ago
5 0
<h3><u>Answer;</u></h3>

Zero

<h3><u>Explanation;</u></h3>
  • <u><em>This is an example of inelastic collision in which energy is lost. </em></u>
  • <u><em>Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy.</em></u>
  • Momentum is conserved, because the total momentum of both objects before and after the collision is the same. However, kinetic energy is not conserved.
  • Kinetic energy is converted into sound, heat, and deformation of the objects.
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Answer:

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<em>Let's take examples to understand. </em>

For example a thread or a table is an object which has a total length of 2 meters.  

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4 0
2 years ago
A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f
Rashid [163]
I believe the answer is #4. u can always ask google if u believe that's the wrong answer :)
4 0
2 years ago
Read 2 more answers
A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so t
Serggg [28]

a) 6.25 rad/s

The law of conservation of angular momentum states that the angular momentum must be conserved.

The angular momentum is given by:

L=I\omega

where

I is the moment of inertia

\omega is the angular speed

Since the angular momentum must be conserved, we can write

L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2

where we have

I_1 = 2.25 kg m^2 is the initial moment of inertia

\omega_1 = 5.00 rad/s is the initial angular speed

I_2 = 2.25 kg m^2 is the final moment of inertia

\omega_2 is the final angular speed

Solving for \omega_2, we find

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular speed

Applying the formula, we have:

- Initial kinetic energy:

K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J

- Final kinetic energy:

K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J

7 0
2 years ago
An alpha particle is identical to a(n) _____.
Alexxx [7]
Helium atom,  in other words, it consistis of a particle having four protons and two neutrons.
3 0
2 years ago
Read 2 more answers
At room temperature what is the strength of the electric field in a 12-gauge copper wire (diameter 2.05 mm that is needed to cau
Dima020 [189]
Electric field strength = resistivity of copper x current density
where
p= 1.72 x 10^-8 <span>ohm meter
diameter = 2.05mm=.00205 m
current = 2.75 A
</span>get first the current density:
current density = current/ cross section area
find the cross section area
cross section area = pi.(d/2)^2;  
cross section = 3.3 006x10-6 m^2
substitute the values 
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current density=35.55 x1 0^2 A/m^2
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Electric field stregnth= 46.415 Volts/m

The electric field strength of copper is 46.415 V/m.


6 0
2 years ago
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