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2 years ago

5

Calculate the speed of an object that moves 200 m in 4 s?

1 answer:

Rasek [7]2 years ago

7 0

50m/s

I divided the 200m by 4s (so 200/4) to get 50m per second

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A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit

Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

$\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E$

now from above equation

$P = 4390 N/m^2$

$\rho = 0.999 * 10^3$

$v = 7.22 * 10^{-4} m/s$

$H = 23.6 m$

now by above equation

$\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E$

$E = 235.7 J/kg$

Part B)

Now energy per unit weight

$U = \frac{E}{g}$

$U = \frac{235.7}{9.8}$

$U = 24 m$

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3 years ago

10. As a spaceship is moving toward Earth, an Earthling measures its length to be 325 m, while the captain on board radios that

Oliga [24]

Answer:

Explanation:

10. As a spaceship is moving toward Earth, an Earthling measures its length to be 325 m, while the captain on board radios that her spaceship's length is 1150 m. (c = 3.00 × 108 m/s) (a) How fast is the rocket moving relative to Earth? (b) What is the TOTAL energy of a 75.0-kg crewman as measured by (i) the captain in the rocket and (ii) the Earthling?

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1 year ago

A bird is about 6.26.2 in. long, with a thin, dark bill and a wide, white wing stripe. If the bird can fly 9292 mi with the w

Trava [24]

Answer:

$209$ mph

Explanation:

$V$ = Speed of bird in still air

$v$ = Speed of wind = 44 mph

Consider the motion of the bird with the wind

$D_{1}$ = distance traveled with the wind = 9292 mi

$t_{1}$ = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

$t_{1} = \frac{D_{1}}{V + v}$

$t_{1} = \frac{9292}{V + 44}$ eq-1

Consider the motion of the bird with the wind

$D_{2}$ = distance traveled against the wind = 6060 mi

$t_{2}$ = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

$t_{2} = \frac{D_{2}}{V + v}$

$t_{2} = \frac{6060}{V - 44}$ eq-2

As per the question,

Time taken with the wind = Time taken against the wind

$t_{1} = t_{2}$

$\frac{9292}{V + 44} = \frac{6060}{V - 44}$

$(9292) (V - 44) = (6060) (V + 44)$

$9292V - 408848 = 6060V + 266640$

$3232V = 675488$

$V = 209$ mph

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2 years ago

A worker uses a cart to move a load of bricks a distance of 10m across a parking lot. if he pushes the cart with a constant forc

andriy [413]

Answer:

2090 J

Explanation:

the work done to move the cart is equal to the product between the force applied and the distance traveled:

$W=Fd$

In this case, the force applied is F=209 N, while the distance covered is d=10 m, therefore the work done is

$W=Fd=(209 N)(10 m)=2090 J$

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2 years ago

Read 2 more answers

Helllllllllllllllllp plss

andrey2020 [161]

Answer:

I don't know answer. ssza

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2 years ago