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k0ka [10]
3 years ago
6

Determine the answer to the equation 30 km/h × 17 h =

Physics
1 answer:
Jobisdone [24]3 years ago
5 0
30 km/h * 17 h =  30*17  km/h  *h
                         =    510 km
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Explain the difference between si base units and derived units. give an example of each
34kurt
Base SI unit is the unit that used for simple quantity like time=second and length= meter. They only have one unit.
Derived unit is more complex because you multiply or divide at least two base SI, making it have more than 1 unit. The example could be velocity which was time/length = m/s
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1.What is an object that appears fuzzy through a material that is?
Anna007 [38]

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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
how does spatial pattern of heights illustrate the relationship between temperature density and the rate of vertical pressure ch
Anika [276]

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Generally, the relationship between temperature, density and rate of vertical pressure is given as;

\rho = \frac{PM}{RT}

\frac{dP}{dz} = -\rho g\\\\

where;

  • <em>ρ is density</em>
  • <em>T is temperature</em>
  • <em>dP is rate of change of vertical  pressure</em>

Thus, from the formula above, we can conclude the following relationship between temperature, density and the rate of vertical pressure change in spatial pattern of heights.

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Learn  more here:brainly.com/question/25395377

5 0
3 years ago
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