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olga nikolaevna [1]
3 years ago
12

As air is heated, the particles speed up and spread out (becoming less dense). Hot air balloon pilots can change the balloon’s a

ltitude (up or down) by turning a heater at the base of the balloon on or off. - If you were piloting a hot air balloon, do you think you would turn the heater ON or OFF to make the balloon go up in the air? WHY do you think that?

Physics
1 answer:
Jobisdone [24]3 years ago
3 0

Answer:

Turn the heater on  

Explanation:

There are two main forces involved in a balloon flight

The downward force is the total weight of the balloon: the air it contains, the gas bag, the basket, the passengers, etc.

The upward force is the weight of the of the air the balloon displaces.

During level flight ,

buoyant force = weight of displaced air - total weight of balloon

If you increase the temperature of the air in the bag, the air molecules spread out and leave through the bottom of the bag.  

The balloon still has the same volume, so the weight of displaced outside air stays the same.

However, the balloon has lost some hot inside air, so its total weight decreases.  

The upward force is greater than the downward force, so the balloon rises.

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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
If the period of the pendulum is tripled, what is the length of the string increased by?
Nitella [24]
The period of the pendulum doesn't determine the length of the string. 
It's the other way around.

The period of the pendulum is proportional to the square root of its length.
So if you want to triple the period, you have to make the string nine times
as long as it is now.
7 0
3 years ago
Why does a dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s?
andrew11 [14]

A dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.

s = vt - 1 / 2 at²

s = Displacement

v = Final velocity

t = Time

a = Acceleration

s = 5 m

t = 1 s

a = 10 m / s²

5 = ( v * 1 ) - ( 1 / 2 * 10 * 1 * 1 )

5 = v - 5

v = 10 m / s

The equation used to solve the given problem is an equation of motion. In a free fall motion, usually air resistance is not considered for easier calculation. If air resistance is considered acceleration cannot be constant throughout the entire motion.

Therefore, a dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.

To know more about equation of motion

brainly.com/question/5955789

#SPJ1

4 0
8 months ago
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
2 years ago
What happens to gas that is not used
kvv77 [185]
the engine won't start or it sputters when it should be running perfectly. if the gasoline is old and stale, it will have lost a portion of its volatility. the lighter components of the gasoline (remember, gasoline is a mixture of different hydrocarbons) have probably evaporated off or disappeared.
3 0
2 years ago
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