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Gemiola [76]
3 years ago
10

How can we prevent ear-problems often encountered by passengers after a flight journey?​

Physics
1 answer:
slamgirl [31]3 years ago
3 0

<u>Follow these tips to prevent ear-problems often encountered by passengers after a flight journey:-</u>

Yawn and swallow during ascent and descent. ...

Use the Valsalva maneuver during ascent and descent. ...

Don't sleep during takeoffs and landings. ...

Reconsider travel plans. ...

Use an over-the-counter nasal spray. ...

Use decongestant pills cautiously. ...

Take allergy medication.

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Valerie is preparing a blood sample for analysis. She wants to separate the solid red blood cells and obtain only the liquid por
liraira [26]
<span>First, she should put the sample in a test tube and place it in a centrifuge. This would cause the red blood cells to move to the bottom because of their higher density. Next, she would be able to decant the plasma and analyze it separately from the red blood cells.</span>
4 0
4 years ago
Read 2 more answers
A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w
gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

Speed of car 1, u₁ = 2 m/s

Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}  

V = 0.46 m/s

So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.          

3 0
3 years ago
Earth is slightly closer to the sun in january than in july. How does the area swept out by earth’s orbit around the sun during
mr Goodwill [35]

This is because both areas are the same.

6 0
3 years ago
Suppose that a nascar race car is moving to the right with a constant velocity of +86 m/s. What is the average acceleration of t
Stels [109]

(a) As the car is moving with constant velocity, it means the rate change of velocity does not change, therefore the average acceleration of the car is zero.

Thus, there is no acceleration, when velocity is constant.

(b) Average acceleration,

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

Here, v is final velocity and u is the initial velocity and \Delta t is the time interval.

As twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed, therefore

a =\frac{ -86 \ m/s-86 \ m/s }{12 \ s} = - 14 .3 \ m/s^2

Thus, the average acceleration of the car is 14 .3 \ m/s^2 in the direction to the left.

6 0
3 years ago
A charged particle is placed in an external magnetic field and it is moving in a circular path of radius 26m.The magnetic force
ElenaW [278]

Answer:

208 Joules

Explanation:

The radius of the circular path the charge moves, r = 26 m

The magnetic force acting on the charge particle, F = 16 N

Centripetal force, F_c = m·v²/r

Kinetic energy, K.E. = (1/2)·m·v²

Where;

m = The mass of the charged particle

v = The velocity of the charged particle

r = The radius of the path of the charged particle

Whereby the magnetic force acting on the charge particle = The centripetal force, we have;

F = F_c = m·v²/r = 16 N

(1/2) × r × F_c = (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.

∴ (1/2) × r × F_c = (1/2) × 26 m × 16 N =  = (1/2)·m·v² = K.E.

∴ 208 Joules = K.E.

The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.

4 0
3 years ago
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