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cupoosta [38]
4 years ago
10

Predict the number of valence electrons for each element based on its location in the periodic table of elements.

Physics
1 answer:
o-na [289]4 years ago
6 0

Answer:

Barium=2

Lead=4

Bismuth=5

Potassium=1

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We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

F_D = c_D A \frac{\rho V^2}{2}

Where,

F_D = Drag Force

c_D = Drag coefficient

A = Area

\rho= Density

V = Velocity

Our values are given by,

c_D = 0.5 (That is proper of a cone-shape)

A = 9m^2

\rho = 1.2Kg/m^3

V = 6.5m/s

Part A ) Replacing our values,

F_D = 0.5*9*\frac{1.2*6.5^2}{2}

F_D = 114.075N

Part B ) To find the torque we apply the equation as follow,

\tau = F*d

\tau = (114.075N)(7)

\tau = 798.525N.m

3 0
4 years ago
What is inside of a blackhole ?
horsena [70]
We don't know. A black hole is a star that has collapsed into its own gravity. The gravity in fact, is so strong that even light cannot get through it. That's why it looks black to us.
7 0
3 years ago
At which point does a substance have the greatest amount of kinetic energy
kogti [31]
The hottest temp i think
8 0
2 years ago
A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is n
ioda

Answer:

13.5

Explanation:

Mass: 5kg

Initial Velocity: -15

Final Velocity: 12

Force: 10

We can use the equation: Vf = Vi + at

We need to find acceleration, and we can use the equation, F=ma,

We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.

Now we have all the variables to find time.

Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)

Plugging them in into desmos gives 13.5 for time.

4 0
2 years ago
Find the resultant of two forces of 4.0N and 6.0N acting at an angle of 160° to each other.
hichkok12 [17]

\star\sf \overrightarrow{A}=4N

\star\sf\overrightarrow{B}=6N

Angle=160

\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}

\\ \sf\longmapsto R=\sqrt{4^2+6^2+2(4)(6)cos160}

\\ \sf\longmapsto R=\sqrt{16+36+48(-0.93)}

\\ \sf\longmapsto R=\sqrt{52-44.6}

\\ \sf\longmapsto R=\sqrt{7.4}

\\ \sf\longmapsto R=2.7N

6 0
3 years ago
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