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Dvinal [7]
2 years ago
9

PLZZZZZ HELPPPPPP MEEEEEEE!!!!!!!! :'(((((

Physics
1 answer:
goblinko [34]2 years ago
8 0

Answer: I'm not 100% sure but I think the answer is A

Explanation: Both fission and fusion are nuclear reactions that produce energy, but the applications are not the same. Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, and fusion is the process where two light nuclei combine together releasing vast amounts of energy.

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(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
3 years ago
Which of the following statements best explains why a book resting on a table is in equilibrium?
timofeeve [1]
I'm pretty sure the answer is d.The weight of the book and the table's upward force on the book are equal in magnitude but opposite in direction.

3 0
3 years ago
Can you be a teen at the age of 12
mr Goodwill [35]

Answer:

12 is a tween (preteen) and still growing, definitely still a kid and not really a teen

4 0
2 years ago
Read 2 more answers
The least common fossils are those that have been<br> petrified<br> frozen<br> buried<br> distilled
Alchen [17]

Answer: Frozen fossils

6 0
3 years ago
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci
MA_775_DIABLO [31]

Answer:

v_B=3.78\times 10^5\ m/s

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is 6.63\times 10^{-27}\ kg

Speed of nucleus at A is v_A=6.2\times 10^5\ m/s

Potential at point A, V_A=1.5\times 10^3\ V

Potential at point B, V_B=4\times 10^3\ V

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s

So, the speed at point B is 3.78\times 10^5\ m/s.

7 0
3 years ago
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