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Klio2033 [76]
3 years ago
14

Which pair below describes isotopes of the same element? A) an atom with 6 protons and 8 neutrons - an atom with 8 protons and 6

neutrons B) an atom with 6 protons and 6 neutrons - an atom with 6 protons and 7 neutrons C) an atom with 8 protons and 8 neutrons - an atom with 7 protons and 8 neutrons D) an atom with 7 protons and 6 neutrons - an atom with 6 protons and 6 neutrons
Physics
2 answers:
iogann1982 [59]3 years ago
7 0

An isotope is an atom with a different number of neutrons than another atom of the same element. Since atoms of the same element all have the same number of protons, choice B(6pro &6neu vs. 6pro&7neu) is an example of isotopes



Otrada [13]3 years ago
4 0

Answer: B)an atom with 6 protons and 6 neutrons - an atom with 6 protons and 7 neutrons

Explanation:

Isotopes are elements which have same atomic number but different mass number.

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

Atomic number is defined as the number of protons or number of electrons that are present in an atom.

Atomic number = Number of electrons = Number of protons

Thus isotopes of same element must have same number of protons and different number of neutrons.

The only combination that fulfil this criteria is : an atom with 6 protons and 6 neutrons - an atom with 6 protons and 7 neutrons

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A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be
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Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t  :

Equations of the uniformly accelerated rectilinear motion of upward   (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m  

y: vertical position in meters (m)  

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j  ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the  initial  vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 =  (v₀y)²

v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }

v₀y = 14.3 m/s

Calculation of the maximum height  the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball  (R)

We replace data in the equation (1)

x =vx*t    vx= 7.7 m/s , t =2.92 s  (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

v= \sqrt{v_{x}^{2}+v_{y}^{2}  }

v= \sqrt{(7.7)^{2}+ (-14.3)^{2}  }

v= 16,2 m/s

\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })

\alpha = tan^{-1} (\frac{-14.3 }{7.7 })

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

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