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sergij07 [2.7K]

2 years ago

5

A. Decide whether the given examples illustrate a constant or a variable.

1 answer:

notka56 [123]2 years ago

5 0

Answer:

a constant is a data item whose value cannot change during the program execution just as its name implies that the value is constant a variable is a data item whose value can change during the program's execution . Thus as its name implies the the value can vary

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gayaneshka [121]

Answer:

10.45?

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2 years ago

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During a construction project, heavy rain filled construction cones with water. The diameter of a cone is 12 in. and the height

sergey [27]

Volume of a cone = <span>πr^2(h/3)
r=6
h=27

v = </span><span>π(6^2)27/3
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3 years ago

IgorC [24]

The answer is (C) the middle one

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3 years ago

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tekilochka [14]

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D

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2 years ago

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John

insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

$H_{max}=\frac{v^2}{2g}$

$H_{max} = \frac{(48)^2}{2 \times 12}$

= 96 ft

b). Time he can stay in the air before hitting the ground is

$T=\frac{2v}{g}$

$T=\frac{2 \times 48}{12}$

= 8 seconds

c). Considering upward motion as positive direction.

v = u + at

We find the time taken to reach the maximum height by taking v = 0.

v = u + at

0 = 16 + (12) t

$t=\frac{16}{12}$

$=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t = $=\frac{4}{3} \ s$ , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$ feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s

3 0

3 years ago