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Hitman42 [59]
2 years ago
14

A cat weighing 7 kg chases a mouse at a speed of 4 m/s. What is the kinetic energy of the cat?

Physics
1 answer:
jeka57 [31]2 years ago
8 0
54J
You calculate kinetic energy by the equation.
KE= 1/2mv^2
m= mass
v= velocity
then we have
KE = 1/2 x 3 x 6^2
KE = 1/2 x 3 x 36 = 54J
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3 years ago
A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin
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Answer:

(A) 0.63 J  

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + mh^{2}  

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = (\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2})^{2}[/tex]

I = (\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2})^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5Iω^{2}

  = 0.5 x 0.07875 x 4^{2} = 0.63 J   0.15 m

(B) from the conservation of energy

   initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

   Ki + Ui = Kf + Uf

   at the maximum height velocity = 0 therefore final kinetic energy = 0

   Ki + Ui = Uf

   Ki = Uf - Ui

 Ki =  mg(H-h)

where (H-h) = rise in the center of mass

     0.63 = 0.42 x 9.8 x (H-h)

   (H-h) = 0.15 m

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Explanation:

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Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small
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Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

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In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

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Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

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