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zvonat [6]
3 years ago
14

How many electrons do the elements of the most stable group have in their outer energy levels

Physics
2 answers:
Wittaler [7]3 years ago
6 0
I think it might be 8 electrons
jekas [21]3 years ago
6 0
The answer is 18 electrons
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2. Sally put her mug of coffee on the table. The mug has a mass of 1 kg. if the pressure
egoroff_w [7]

We have,

  • The mass of sally's mug is 1 kg
  • The pressure appliedby the mug is 1100 pascal.

We know that,

  • Pressure = Force/Area

As, we already have the value of pressure, let's calculate that of force now;

  • F = ma
  • F = 1 × 9.8 { Acceleration due to gravity, let's round off it to 10}
  • F = 10 N

Just put all the values in the formula now;

  • P = F/A
  • 1100 = 10 / A
  • 1100/10 = A
  • 110 m² = A

As, it is already mention that we need to find the radius of the mug, it is obviously a circular base.

We know that,

  • Surface area = Circumference

So, let's solve it;

  • Circumference = 2πr
  • 110 = 2 × 22/7 × r
  • 110 × 7/2 × 22 = r
  • 5 × 7 = r
  • 35 cm = r

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3 years ago
A baseball is hit at an initial speed of 40 m/s at an angle of 60° above the horizontal and reaches a maximum height of h meters
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C) 6h i’m not that great at math but that’s what i got hope it helps!!
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A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first
Dominik [7]

Answer:

a) V=14.01 m/s

b) V=8.86 \, m/s

c)t = 2.33s

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

E_m = E_k +E_p

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this E_m mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

E_k = \frac{1}{2} m \, V^2

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

E_p= m\, g\, h

where g is the acceleration due to gravity (g=9.81 \, m/s^2) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which h=0, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set h=0

E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\

And so we have found the velocity of the ball as it hits the floor.

V = \sqrt[]{2g\cdot 10m}=14.01\, m/s

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

V = \sqrt[]{2g\cdot h}

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

V_{avg}\cdot t=h\\t=h/V_{avg}

what's the average speed when the ball is descending?

V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s

so the time it takes the ball to go down is:

t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s

and the time it takes to go up is:t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s

When we add both times , we get:

t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s

6 0
3 years ago
In zoes egg experiment which two forces are balanced to keep it from hitting the pan
alexgriva [62]

Answer:

The forces of push and pull

Explanation:

In the egg drop experiment, the egg is balanced on top of a toilet paper tube and balanced on a pan. The pan itself is placed on top of a glass of water. When the experimenter uses his hands to push the pan, the force of gravity pulls the egg downwards making it move down, right into the glass of water.

So the two forces which are applied in opposite directions are;

1. The force of push applied horizontally by the hand when it pushes the pan away, and

2. The force of pull caused by gravity which makes the egg move vertically downwards.

3 0
3 years ago
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