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Leno4ka [110]
3 years ago
5

WILL GIVE BRAINLIEST!

Physics
2 answers:
ASHA 777 [7]3 years ago
5 0

Answer:

A: sublimation

B: vaporization

C: melting

Explanation:

Bad White [126]3 years ago
3 0

Answer:

A. Deposition

B. Condensation

C. Freezing

Explanation:

We have three changes of state shown in the picture:

- Deposition: it is the change of state of a substance from gas to solid state, without going through the liquid phase. In this process, energy is released, as the average energy of the molecules of the substance decrease, releasing it to the surroundings.

- Condensation:  it is the change of state of a substance from gas to liquid state. In this process, energy is also released, as the average energy of the molecules of the substance decrease as the substance becomes a liquid, releasing it to the surroundings.

- Freezing: it is the change of state of a substance from liquid to solid state. In this process, energy is also released, as the average energy of the molecules of the substance decrease as the substance becomes a solid, releasing it to the surroundings.

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Answer:

B) Because the Space Station is constantly in free-fall around the Earth.

Explanation:

Anything that is falling experiences an upward force on them. For example when a person is going down in a lift they will experience something that is pushing them upwards. This happens due to the fact that the total acceleration the body is feeling is less than the acceleration due to graviity.

The force on a body which is falling is

F=m(g-a)

Where,

m = Mass of object

g = acceleration due to gravity

a = acceleration the object is experiencing.

a = g. So, the force becomes zero and the object experiences weightlessness.

Hence, the astronauts in the space station experience weightlessness due to fact that the Space Station is constantly in free-fall around the Earth.

7 0
3 years ago
Difference between relaxation time and collision time?
Black_prince [1.1K]
For the answer to the question above, let us first start with relaxation time. it is the absence of an external electric field, the free electrons in a metallic substance will move in random directions so that the resultant velocity of free electrons in any direction is equal to zero. While the Collision time it is<span> the mean </span>time<span> required for the direction of motion of an individual type particle to deviate through approximately as a consequence of </span>collisions<span> with particles of type.</span>
6 0
3 years ago
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What is one disadvantage of a series circuit?
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6 0
3 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
3 years ago
Which body is in equilibrium?
sergey [27]
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