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omeli [17]
3 years ago
14

What is the differ of fermentation from respirstion

Physics
1 answer:
Andrej [43]3 years ago
8 0
This may help u... :P

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A beam of helium-3 atoms (m = 3.016 u) is incident on a target of nitrogen-14 atoms (m = 14.003 u) at rest. During the collision
Montano1993 [528]

Answer:

Explanation:

We shall apply law of conservation of momentum to solve the problem .

Helium-3 collides with nitrogen-14 at rest . After the collision the newly formed deuterium atom and oxygen-15  atom moves .

momentum before the collision

= 3.016 x 6.346 x 10⁶ + 14.003 x 0 = 19.14 x 10⁶ unit

momentum after collision

2.014 x 1.531 x 10⁷ + 15.003 V

3.083 x 10⁷ +  15.003 V units

Applying the law of conservation of momentum ,

19.14 x 10⁶ = 3.083 x 10⁷ +  15.003 V

1.914 x 10⁷ = 3.083 x 10⁷ +  15.003 V

15.003 V = - 1.169 x 10⁷

V = .077917 x 10⁷

= 7.79 x 10⁵  m /s

= .0779 x 10⁷ m /s

mass of helium atom = 3.016 u = 3.016 x 1.67 x 10⁻²⁷ kg

velocity = 6.346 x 10⁶ m /s

kinetic energy = 1 /2 x  3.016 x 1.67 x 10⁻²⁷ x (6.346 x 10⁶ )²

= 101.42 x 10⁻¹⁵ J  

kinetic energy of nitrogen atoms = 0

Total energy before collision =  101.42 x 10⁻¹⁵ J  

Similarly kinetic energy after collision

= 1 /2 x [ 2.014 x 1.531² + 15.003 x .0779² ] x 1.67 x 10⁻²⁷ x 10¹⁴

= .835 x [ 4.72  + .09 ] x 10⁻¹³ J

=  4.016 x 10⁻¹³ J

= 401.6 x 10⁻¹⁵  J  

value of kinetic energy is increased .

5 0
3 years ago
The 20 oz orange soda you drank at lunch contained 1 oz. Of real orange juice. What percent of the orange soda is real orange ju
Dimas [21]
Answer: 5%

You divide 1 by 20

You will get 0.05

Move the decimal 2 digits to the right

=5
6 0
3 years ago
Can u help me pls....
Schach [20]

Answer:

a) m = 7000 - 500 = 6500 g

b) ρ = 6500 g / 5000 cm³ = 1.3 g/cm³

c) Brush floats because its density is less than that of the paint and displaces its own weight in the paint.

6 0
3 years ago
Physics
mixas84 [53]

Answer:

3 and 2

May be

ok may be subscribeIsenberg Isenberg Ignatius

3 0
2 years ago
I shared a picture of the problem. It’s a basic Physics question and an Algebra question.
julia-pushkina [17]

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

Given the expressions;

T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}

Equating both expressions we will have;

2 \pi \sqrt{\frac{m}{k} }  = \frac{2 \pi}{\omega}

Divide both equations by 2π

\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\

Square both sides

(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}

Take the square root of both sides

\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

3 0
3 years ago
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