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3 years ago

Hii I need help can someone help me

Engineering

1 answer:

zavuch27 [327]3 years ago

3 0

Answer:

I'm not 100 percent sure, but I'm pretty sure that it is half of an ellipse.

Explanation:

For one, if you look at a picture of a hyperbola, you can see that it looks like nothing and is nothing compared to the parabola. However, looking at an ellipse, you can clearly see the resemblance between the two.

Once again, I am not 100 percent certain with this answer, so if you have anything to say about it please let me know.

Have a nice day!

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Multiple Choice

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6 0

3 years ago

A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a

dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have $volume of box = 25ft*10ft*12ft= 3000ft^{3}$

$Volume= 84.95m^{3}$

$Since 1ft^{3} =0.028m^{3}$

Now the weight of water displaced = $Weight =\rho \times Volumewhererho$ is density of water = 1000kg/ $m^{3}$

Thus weight of liquid displaced = $\frac{84.95X1000}{1000}tonnes=84.95 tonnes$ ..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

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3 years ago

A machine raises 20kg of water through a height of 50m in 10secs. What is the power of the machine.

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Answer:

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Explanation:

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2 years ago

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

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Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

$\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)$

solvingt for $\epsilon)$

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3 years ago

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Answer:

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Explanation:

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2 years ago