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3 years ago

Ex1. An atom has atomic number Z = 6 and mass number A = 12. Establish:

Physics

1 answer:

Maksim231197 [3]3 years ago

7 0

Answer:

Ex1.

a) Carbon : atomic number Z = 6 and mass number A = 12

b) Protons = 6; electrons = 6 ; neutrons = 6.

Ex2.

Molar mass of H₂O₂ = 34 g; BaF = 156 g; CO = 28 g.

Explanation:

Ex1.

a) the element with atomic number 6 and mass number 12 is carbon

b) number of protons and electrons in a element is given by its atomic number; and for neutrons its mass number - atomic number (A-Z).

⇒ protons = Z = 6; electrons = Z = 6.; neutrons = A-Z = 12-6 = 6.

Ex2.

Molecular formula of hydrogen peroxide is given by H₂O₂.

⇒ molar mass = 2×mass of hydrogen + 2×mass of oxygen

⇒ molar mass = 2×1 + 2×16 = 34.g

molar mass of BaF = mass of barium + mass of fluoride.

= 137 + 19 = 156 g

molar mass of CO = mass of carbon + mass of oxygen

= 12 + 16 = 28 g

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Answer:

(a)1.557 radian (b) 1.424 A

Explanation:

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So phase difference $=\frac{2\times 3.14\times 0.85\times 100}{343}=1.557radian$

(b) The resultant amplitude is given by $2Acos\frac{\Phi }{2}=2\times A\times cos\frac{1.557}{2}=1.424A$

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2 years ago

During the launch from a board, a diver's angular speed about her center of mass changes from zero to 5.10 rad/s in 170 ms. Her

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Answer:

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3 years ago

the time required for one cycle, a complete motion that returns to its starting point, it called the_____ medium frequency perio

telo118 [61]

Answer:

The correct answer to the following question will be "Period".

Explanation:

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3 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

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t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

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Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

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Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

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2 years ago

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Scorpion4ik [409]

Answer:

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L = 0.151027... m

L = 15 cm

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2 years ago

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