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lutik1710 [3]
3 years ago
13

A hurdler is 0.535 m from a hurdle when he jumps at 6.82 m/s at a 6.79 degree angle. What is his height when he clears the hurdl

e?(Hint: It is not necessarily at the highest point in his motion.) Unit = m. Would love an explanation for this physics problem!
Engineering
1 answer:
butalik [34]3 years ago
3 0

Answer:

0.82 m

Explanation:

have a good day

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A gear train has two gears. The driver gear has 8 teeth and a diametral pitch of 6 teeth/inch. the follower gear has 24 teeth. W
Sliva [168]

Answer:

18 teeth/inch

Explanation:

Given that: i. driver gear has 8 teeth and diametral pitch of 6 teeth/inch.

                  ii. follower gear has 24 teeth.

Let the followers diametral pitch be represented by x.

Then,

8 teeth ⇒  6 teeth/inch

24 teeth ⇒ x teeth/inch

So that;

x = \frac{24 x 6}{8}

   = \frac{144}{8}

   = 18 teeth/inch

The diametral is 18 teeth/inch

3 0
3 years ago
If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo
Stolb23 [73]

Answer:

526.5 KN

Explanation:

The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.

But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.

h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg

where ρ = density of the fluid and g = acceleration due to gravity

h = ΔP/ρg

ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa

Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with

Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa

Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²

Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN

3 0
3 years ago
Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
Zina [86]

Answer:

0.245 m^3/s

Explanation:

Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s

5 0
3 years ago
Just some random stufff
NeTakaya

Answer:

Nice!

Explanation:

5 0
3 years ago
Which of the following best distinguishes between superficial design improvements and functional
Contact [7]

Answer:

Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.

Explanation:

As a PC board designer, I would sometimes spend a certain amount of time making traces have shorter routes, or fewer layer changes or bends. (I wanted to make the layout "pretty.") In some cases, these changes are superficial, affecting the appearance only. In some cases, they are functional, reducing crosstalk or emissions or susceptibility to interference.

I deal with a web site that seems to be changing all the time (Brainly). In many cases, the same information is rearranged on the page—a superficial change. In other cases, the information being displayed changes, or the way that certain information is accessed changes. These are functional changes. (Sometimes, they "enhance performance," and sometimes they don't, IMO.)

In short ...

<em>Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.</em>

8 0
2 years ago
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