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3 years ago

6

Which action best demonstrates the transformation of mechanical energy to heat energy? Select one: a. Digesting a meal b. Burnin

g a candle c. Rubbing hands vigorously d. Melting a piece of ice in sunlight

1 answer:

Anton [14]3 years ago

4 0

<em>(c). </em>Rub vigorously (mechanical) ===> hands get hot (heat energy).

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Based on the diagram which of the following statements is true of a helium atom

snow_lady [41]

Answer: Option 2

Explanation:

Option 1 is wrong because there are 2 protons and 2 neutrons in nucleus.

Option 3 is wrong because there are two electrons moving around nucleus.

Option 4 is wrong because electrons are negatively charged and are moving around the nucleus.

Option 5 is wrong because there equal amount of protons and electrons with 2 each.

8 0

3 years ago

Considering the various theories, the energy used in forming organic molecules in the primitive atmosphere could have come from

OLEGan [10]

Answer:

<h2>e. sound. </h2>

Explanation:

Such type of atmosphere in which oxygen was not present or was present in little amount is known as the primitive atmosphere and such type of atmosphere was present in the initial stage of the earth formation.
During this period, water vapor, nitrogen, hydrogen and carbon dioxide gases were present.
These gases interact with the help of energy that comes from many sources such as lightning, ultraviolet radiation, electric spark and some other.
When these inorganic molecules react in the then the formation of organic compounds takes place that becomes the basis of the organization of the life on the earth and called an organic evolution of life.

8 0

2 years ago

What are three observations an astronaut might make while viewing Russia at night

Lunna [17]

1. Physical size of Russia compared to other countries, despite a lack of visible borders from space.

2. Part of Russia's outline would likely be obscured by the various clouds and objects in the stratosphere; this would allow the astronaut to view potential cloud and weather patterns on earth.

3. An astronaut could see outlines of Russia's geography such as mountain ranges.

Hope that it helps :)

7 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

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0

3 years ago