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Lubov Fominskaja [6]
3 years ago
7

What is the velocity of a jet that travels 785 m [W] in 2.39 s?

Physics
1 answer:
Cerrena [4.2K]3 years ago
7 0

Answer:

v = 328.4

Explanation:

Data:

  • Distance (d) = 785 m
  • Time (t) = 2.39 s
  • Velocity (v) = ?

Use formula:

  • \boxed{v = \frac{d}{t}}

Replace:

  • \boxed{v = \frac{785m}{2.39s}}

It divides:

  • \boxed{v = 328.4\frac{m}{s}}}

What is the velocity?

The velocity is <u>328.4 meters per second.</u>

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If the architectural plans show the rough opening of a window to be 3'-3" x 4'-9" , the height of the opening should actually me
tatyana61 [14]

Answer:

height of the opening actually measure is 4'-9"

Explanation:

given data

window size = 3'-3" x 4'-9"

solution

height of the opening should actually measure will be 4'-9" in 3'-3" x 4'-9"

because according to architectural plan height can not be more than the opening size of window

and we can't take smaller height also

so fit in opening window we should take same height of  height of opening window and that is here 4'-9"

so here height of the opening actually measure is 4'-9"

5 0
3 years ago
A(n) 60.3 g ball is dropped from a height of 53.7 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Feliz [49]

Answer:

271.862 N/m

Explanation:

From Hook's Law,

mgh = 1/2ke²............... Equation 1

Where

m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.

Making k the subject of the equation,

k =2mgh/k²....................... Equation 2

Note: The potential energy of the ball is equal to the elastic potential energy of the spring.

Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m

Substitute into equation 2

k = 2(0.0603)(9.8)(0.537)/0.048317²

k = 0.6346696/0.0023345

k = 271.862 N/m

7 0
3 years ago
Water is a fluid, all fluids
Andreyy89
What is the question?
4 0
4 years ago
A car goes from 5 m/s to 25 m/s in 6 s. What is the acceleration of the car?
damaskus [11]

Answer:

5m/s

Explanation:

3 0
3 years ago
The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0
4 years ago
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