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Levart [38]
3 years ago
11

A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy​

Physics
1 answer:
AURORKA [14]3 years ago
8 0

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

_________________________________

Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

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Ernest Rutherford (who was the first person born in New Zealand to be awarded the Nobel Prize in chemistry) demonstrated that nu
g100num [7]

Answer:

the velocity of the gold nuclei  v'_2 = 5.34 *10^5 \ m/s

the angle of the velocity of gold nuclei = 29.64° clockwise (i.e down the horizontal)

the final kinetic energy of helium nucleus is 7.52 *10^{-13} \ \ J

Explanation:

The formula for kinetic Energy (K.E) is:

K.E = \frac{1}{2}m_1v_1^2

v_1 = (\frac{2K.E}{m_1})^{1/2}

where;

m_1 = mass of the helium

v_1 = velocity of helium

replacing 8.00*10^{-13} \ J \ for \ K.E and m \ with \ 6.68*10^{-27} \ kg; we have :

v_1 = (\frac{2*8.00*10^{-13 \  }J}{6.68*10^{-27} kg})^{1/2}

v_1 = 1.548*10^7  \  m/s

Applying conservation of momentum along x- axis since the collision is elastic

m_1v_1 =m_1v'_1cos \theta _1 + m_2v_2' cos \theta_2         ------ equation (1)

where

m_1 \ and \ m_2 = mass of helium and gold respectively

v_1 = initial velocity of helium

v'_1 \ and  \ v'_2 = final velocity of gold

\theta _1 \ and  \ \theta _2 = scattered angle for helium  and gold respectively

Along the y - axis; the equation for conservation of momentum is :

0 = m_1 v_1' \theta_1 + m_2 v_2' sin \theta _2  ---- equation (2)

Equating equation (1) and (2); we have:

(v'_2)^2  = \frac{m_1^2}{m_2^2}[(v_1)^2 - 2 v_1v_1' xos \theta _1 + (v'_1)^2]    ----- equation (3)

However, the conservation of internal kinetic energy guves:

\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1(v_1')^2 + \frac{1}{2}m_2(v'_2)^2 \frac{m_1}{m_2}

Making (v_2')^2 the subject of the formula ; we have:

(v_2')^2 = \frac{m_1}{m_2}(v_1^2-(v_1')^2)      ----- equation (4)

Replacing the expression of (v'_2)^2 in equation (3) into equation (4) ; we have

[(1+\frac{m_1}{m_2})(v_1')^2-2(\fracm_1}{m_2}(v_1'cos \theta _1)v_1 - (1-\frac{m_1}{m_2})(v_1)^2] = 0

In the above expression;

replacing ;

1.548*10^7 \ m/s for \ v_1;   \theta = 120^0;   m_1 = 6.68*10^{-27} \ kg;  m_2 = 3.29*10^{-25}\ kg; we have:

[1 + ( \frac{6.68*10^{-27}}{3.29*10^{-25}})(v_1')^2 - ( 2(\frac{6.68*10^{-27}}{3.29*10^{-25}}) (1.548*10^7)cos 120^0)v'_1-(1-\frac{6.68*10^{-27}}{3.29*10^{-25}})(1.548*10^7)^2 = 0

1.02v^2 - (3.143 *10^5)v -2.348*10^{14} = 0

The above is a quadratic equation; now solving by using the quadratic formula; we have:

v_1' = \frac{-3.143*10^5 \pm \sqrt{(3.143*10^5)^2 -4(1.02)(-2.348*10^{14})}}{2(1.02)}

since we are considering the positive value from the above expression; we have

v'_1 = 1.50*10^7 \ m/s

NOW; we substitute our known values into equation (4) in order to solve for v_2'; we have:

(v'_2 )^2 = \frac {6.68*10^{23}}{3.29*10^{-23}}((1.548*10^7)^2-(1.502*10^7)^2)

(v'_2 )^2 =2.848*10^{11} \ m^2/s^2

(v'_2 )=\sqrt{2.848*10^{11} \ m^2/s^2}

v'_2 = 5.34 *10^5 \ m/s

Therefore; the velocity of the gold nuclei  v'_2 = 5.34 *10^5 \ m/s

From equation (2)

0 = m_1 v_1' \theta_1 + m_2 v_2' sin \theta _2

Therefore replacing our known values and solving for \theta_2; we have:

sin \theta _2 =\frac{(6.68*10^{-27})(1.50*10^7)sin 120^0}{(3.29*10^{-25}(5.34*10^5)}

sin \theta _2 = -0.4946

\theta _2 =sin^{-1}( -0.4946)

\theta _2 = -29.64^0

∴ the angle of the velocity of gold nuclei = 29.64° clockwise (i.e down the horizontal)

b) Equation to determine the final kinetic energy (K.E_f)of helium is:

K.E_f = \frac{1}{2}m_1(v_1')^2

= \frac{1}{2}(6.68*10^{-27})}(1.50*10^7)^2

= 7.52 *10^{-13} \ \ J

Thus, the final kinetic energy of helium nucleus  is 7.52 *10^{-13} \ \ J

I hope this explanation helps alot!.

5 0
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Marina86 [1]

Answer:

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Explanation:

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8 0
3 years ago
50 points!
Tanya [424]
From Kepler's 3rd law, time period of satellite orbiting around planet in a circular orbit is given by,
T² = 4π²r³/GM 
M= mass of the Earth = 6×10²⁴ Kg. 
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴ 
   T² = 1653339719
∴ T = 40661.28 seconds.
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3 years ago
Math hard for me now <br> Help me please please help I bean do it a lot
wlad13 [49]
5 is a prime number because it only has two factors which are 1 and 5.
4 0
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Deduced hydrochloric acid is a strong acid ​
Zina [86]

Answer:

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Explanation:

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