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3 years ago

If the time for F18 to take off doubles, the acceleration will ____.

Physics

1 answer:

kherson [118]3 years ago

8 0

Answer:

decreases

Explanation:

a=[v-u]/t

as time increases acceleration decreases and vice versa

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The position of an electron is given by , with t in seconds and in meters. At t = 3.99 s, what are (a) the x-component, (b) the

egoroff_w [7]

Answer:

A. Vx = 3.63 m/s

B. Vy = -45.73 m/s

C. |V| = 45.87 m/s

D. θ = -85.46°

Explanation:

Given that position, r, is given as:

r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk

Velocity is the derivative of position, r:

V = dr/dt = 3.63 - 11.46t^j

A. x component of velocity, Vx = 3.63 m/s

B. y component of velocity, Vy = -11.46t

t = 3.99 secs,

Vy = - 11.46 * 3.99 = -45.73 m/s

C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]

|V| = √(2091.2329 + 13.1769)

|V| = √(2104.4098)

|V| = 45.87 m/s

D. Angle of the velocity relative to the x axis, θ is given as:

tanθ = Vy/Vx

tanθ = -45.73/3.63

tanθ = -12.6

θ = -85.46°

7 0

3 years ago

What is a literature review?<br>

almond37 [142]

Answer: A literature review consists of an overview, a summary, and an evaluation (“critique”) of the current state of knowledge about a specific area of research.

Explanation:

7 0

2 years ago

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

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2 years ago

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tresset_1 [31]

The answer is C

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3 years ago

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At which temperature will water boil when the external pressure is 17.5 torr

ch4aika [34]

<span>The temperature of water will boil at one hundred degrees celsius when the external pressure is at 17.5 torr. Essentially, it is based off of the vaporizing of heat, as well as the gas constant. This is a matter of solving a physics equation and breaking down the factors that will affect the boiling point.</span>

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3 years ago

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