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2 years ago

Why do white gassy stuff come out of planes?

Physics

1 answer:

ANTONII [103]2 years ago

6 0

Answer:

<h2>Jets leave white trails, or contrails, in their wakes for the same reason you can sometimes see your breath. The hot, humid exhaust from jet engines mixes with the atmosphere, which at high altitude is of much lower vapor pressure and temperature than the exhaust gas.</h2>

Explanation:

<h2>Hopes this helps. Mark as brainlest plz!</h2>

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An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof

Blizzard [7]

240 meters that's it

7 0

2 years ago

What is (3.25 x 10^5) + (7.5 x 10^4)?

ivanzaharov [21]

Answer:

400000

Explanation:

So first solve one part:

(3.25 * 10^5)

(3.25 * 100,000)

= 325000

Then solve the next part:

(7.5 * 10^4)

(7.5 * 10000)

= 75000

Now lastly, add the two answers:

325000 + 75000 = 400000

Therefore,

(3.25 x 10^5) + (7.5 x 10^4) = 400000

5 0

2 years ago

Read 2 more answers

A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails

harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

$y=h+ut+\frac{1}{2}gt^2$

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

$y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m$

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

$v^2-u^2 = 2ad$

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

$v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s$

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

$a=\frac{v-u}{t}$

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

$t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s$

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0

2 years ago

An object is accelerating at 2m/s^2. What is the net force if the mass is 125kg?

natita [175]

Answer:

250 N

Explanation:

Newton's second law:

∑F = ma

∑F = (125 kg) (2 m/s²)

∑F = 250 N

3 0

2 years ago

Which examples are a COMPLETE description of motion? Select ALL that apply.

SIZIF [17.4K]

All of them apply because they all are doing some thing to move

8 0

2 years ago