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3 years ago

Without oceans, Earth would have what?

Physics

1 answer:

PtichkaEL [24]3 years ago

5 0

Answer: Less rain for sure.

Explanation: Without clouds forming over the ocean, rain would be incredibly rare, and the planet would become desert. So if we were to lose all oceans, less rain and clouds would form and eventually we could run out of water. I hope this helps:)

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Describe a technology used in space exploration.

Alex777 [14]

Answer:

High speed optical communication technology

To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.

Explanation:

Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.

The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.

This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication

3 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

A 6 kg weight is lifted off the ground to a height that gives it 70.56 j of gravitational potential energy. what is its height?

Marina CMI [18]

GPE= 70.56 J -------------------> GPE= mgh-------------> X= height
70.56 = 6(kg) * 9.8(m/s/s) * X
70.56 = 58.8X
70.56/58.8= 58.8X/58.8
X= 1.2
The height is 1.2 feet or meters (whatever unit you are using in this problem)

5 0

3 years ago

Read 2 more answers

Sunlight reflects from a concave piece of broken glass, converging to a point 34 cm from the glass. what is the radius of curvat

wolverine [178]

The rays of light coming from the Sun are parallel to each other, so when they are reflected by the concave piece of glass (which acts as a concave mirror) they converge into the focus of the mirror, which is
$f=34 cm$
The radius of curvature of a concave mirror is twice its focal length, so in this case it is:
$r=2f = 2 \cdot 34 cm=78 cm$

6 0

3 years ago

Explain two scenarios where a large truck can have the same momentum as a small car.

KengaRu [80]

The momentum, p, of any object having mass m and the velocity v is

$p=mv\cdots(i)$

Let $M_L$ and $M_S$ be the masses of the large truck and the car respectively, and $V_L$ and V_S be the velocities of the large truck and the car respectively.

So, by using equation (i),

the momentum of the large truck $= M_LV_L$

and the momentum of the small car $= M_SV_S$ .

If the large truck has the same momentum as a small car, then the condition is

$M_LV_L=M_SV_S\cdots(ii)$

The equation (ii) can be rearranged as

$\frac {M_L}{M_S}=\frac {V_S}{V_L} \; or \; \frac{M_L}{V_S}=\frac{M_S}{V_L}$

So, the first scenario:

$\frac {M_L}{M_S}=\frac {V_S}{V_L}$

$\Rhghtarrow M_L:M_S=V_S:V_L$

So, to have the same momentum, the ratio of mass of truck to the mass of the car must be equal to the ratio of velocity of the car to the velocity of the truck.

The other scenario:

$\frac{M_L}{V_S}=\frac{M_S}{V_L}$

$\Rhghtarrow M_L:V_S= M_S:V_L$

So, to have the same momentum, the ratio of mass of truck to the velocity of the car must be equal to the ratio of mass of the car to the velocity of the truck.

5 0

3 years ago