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bagirrra123 [75]
2 years ago
5

Please help! Kiki makes a table to compare the particles in a magnesium atom to those in a magnesium ion. She knows that a magne

sium ion has an electric field that points away from the ion. What values should she use to complete her table?
(The table looks something like this:
Magnesium Atom  Magnesium Ion
Protons (Magnesium Atom)12  (and Magnesium Ion) X
Neutrons (MA)12  (MI)Y
Electrons (MA) 12  (MI) Z
A. X: 14 Y: 12 Z: 12
B.X: 12 Y: 12 Z: 10
C. X: 12 Y: 10 Z: 12 X: 12
D. Y: 12 Z: 14
Physics
2 answers:
Liono4ka [1.6K]2 years ago
7 0
The answer would be D
Alex2 years ago
5 0

Answer:

B.X: 12 Y: 12 Z: 10  

Explanation:

<u>Electric field points away from a positive ion and towards for a negative ion. An atom becomes an ion when it loses or gains electrons. </u>

A positive ion is formed when the electrons are lost to another atom and remaining atom has more number of protons than electrons.

Atomic number of magnesium is 12 which means it has 12 protons and 12 electrons in a neutral atom. atomic mass of magnesium is 24 which means it has 12 neutrons.

A positive magnesium ion must have electrons less than protons. There would be no change in number of protons and neutrons.

<u>From the given options, in option B, number of protons and neutrons is same i.e. 12 where as number electrons is 10 which is less than the number of protons. This means these are the correct options for magnesium ion. </u>

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A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>
  1. Determine the acceleration of the stuffed bear brainly.com/question/6268248
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Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0
2 years ago
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Explain how the forces need to change so the aeroplane can land
Fofino [41]
When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.
7 0
2 years ago
Scotesia swims from the north end to the south end of a 50.0 m pool in 20.0 s. As she begins to make the return trip , Sean, who
slega [8]

Answer:

a) 2.5m/s

b) 0.91m/s

c) 0m/s

Explanation:

Average velocity can be said to be the ratio of the displacement with respect to time.

Average speed on the other hand is the ratio of distance in relation to time

Thus, to get the average velocity for the first half of the swim

V(average) = displacement of first trip/time taken on the trip

V(average) = 50/20

V(average) = 2.5m/s

Average velocity for the second half of the swim will be calculated in like manner, thus,

V(average) = 50/55

V(average) = 0.91m/s

Average velocity for the round trip will then be

V(average) = 0/75, [50+25]

V(average) = 0m/s

3 0
2 years ago
You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the tr
Nitella [24]

Answer:

51.2 mi/h

Explanation:

Total distance, d = 100 miles

First 60 miles with speed 55 mi/h

Next 40 miles with speed 75 mi/h

Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h

Time taken for 40 miles, t2 = 40 / 75 = 0.533 h

Time spent to get stuck, t3 = 20 min = 0.33 h

Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h

The average speed is defined as the ratio of total distance traveled to the total time taken.

Average speed = =\frac{100}{1.953}=51.2 mi/h

Thus, the average speed of the journey is 51.2 mi/h.

4 0
2 years ago
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How lon
frosja888 [35]

Answer:

t=0.704s

Explanation:

A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?

from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed

V^{2} =u^{2} +2as

Vy^2=0+2*9.8*1.73sin40.5

Vy^2=22.021

Vy=4.69m/s

Vx^2=u^2+2*9.81*cos40.5

Vy^2=25.81

Vy=5.08m/s

V=(Vy^2+Vx^2)^0.5

V=47.71^0.5

V=6.9m/s

from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.

f=force applied

v=velocity final

u=initial velocity

m=mass of the toy, 0.046

f=ma

f=m(v-u)/t

v=u+at

6.9=0+9.8t

t=6.9/9.81

t=0.704s

4 0
2 years ago
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