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Lena [83]
3 years ago
14

Choose two forces and compare and contrast these forces. These must be different forces than used in the prior question. Provide

two ways that they are similar and two ways that they are different. You may make a list, write it out, or make a chart. (5pts) (FLVS assignment 01.04) PLEASE HELP!!!
Physics
1 answer:
marin [14]3 years ago
6 0

Answer:

Explanation:

There are 4 forces. These are 1) Gravity, 2) Weak Nuclear Force, 3) Electromagnetism, and 4) Strong Nuclear Force. 

Order of strength from weakest to strongest: Gravity, Weak Nuclear Force, Electromagnetism, Strong Nuclear Force

Type of Range:

Gravity - Unlimited range

Weak Nuclear Force - Limited range

Electromagnetism - Infinite range

Strong Nuclear Force - Limited Range

Found in:

Gravity - Exists between all objects with mass

Weak Nuclear Force - Governs over beta decays like the emission of electron or positron

Electromagnetism - the attraction found between particles that are electrically charged

Strong Nuclear Force - Found in atoms and subatomic particles. It is responsible for holding the atoms' nucleus together.

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calculate the density of a neutron star with a radius 1.05 x10^4 m, assuming the mass is distributed uniformly. Treat the neutro
Orlov [11]

To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

\rho = \frac{m}{V}

Where

m = Mass (Neutron at this case)

V = Volume

The mass of the neutron star is 1.4times to that of the mass of the sun

The volume of a sphere is determined by the equation

V = \frac{4}{3}\pi R^3

Replacing at the equation we have that

\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}

\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}

\rho = 5.75*10^{17}kg/m^3

Therefore the density of a neutron star is 5.75*10^{17}kg/m^3

4 0
3 years ago
Please help me it is for learning at home.
Temka [501]

Answer:

What do you need help with?

Explanation:

8 0
3 years ago
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Intelligence test. If you get it right ,you are a critical thinker. You were in the garden,there are 34 people in the yard. You
bagirrra123 [75]

The riddle is solved by knowing that there is only one person left in the garden and that is you.

<h3>How many people are in the garden?</h3>

We are told that there were a total of 34 people in the yard. We do not know if these people are in the garden or somewhere around the yard. The next information we have is that a total of 30 persons were killed.

Since you killed them in the garden, the others may still be lurking somewhere in the yard thus there is only one person left in the garden and that is you.

Learn more about solving a riddle:brainly.com/question/25063408

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8 0
2 years ago
(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat
vitfil [10]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i)  the gradient of f =  ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is  ñ∙∇F =  4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+  ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+  (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have  

∇f(x, y) = -1·i -3·j  that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.  

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z)  = ∇F(x, y, z) = = ∂f/∂x i+  ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore  

ñ = ⅟4(2·i +3·j -√3·k)  

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)  

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

8 0
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