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

slamgirl [31]

<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.

3 0

3 years ago

Mass m is fired to the right at a speed of 7.0 m/s where it collides and becomes imbedded in mass M which is hanging from a rope

Tanya [424]

Answer:

V=0.28m/s, the combined mass will be 0.28m/s fast

h=0.00399m

Explanation:

Mass m is fired to the right at a speed of 7.0 m/s where it collides and becomes imbedded in mass M which is hanging from a rope. Mass m is 0.50kg and mass M is 12.0kg. a- How fast will the combined mass (m+M) be moving just after the collision? b- What is the maximum height the system (m+M) will reach?

step by step explanation:

from the principles of linear momentum which states that the total momentum before collision is equal to the total momentum after collision

momentum before collision=momentum after collision

notice that in this this case the collision is inelastic, owing to the fact that the two bodies stuck together after collision

mu+Mu2=(m+M)V

V=common velocity after collision

m=mass of the body fired

M=Mass hanging on the rope

u=velocity of the body

u2=velocity of the mass hanging on arope 0m/s

0.5*7+0=(0.5+12)V

3.5=12.5V

V=3.5/12.5

V=0.28m/s, the combined mass will be 0.28m/s fast

b. the maximum height reached after the impact

starting with the swing of the pendulum just after the collision until it reaches its maximum height. Conservation of mechanical energy applies here, so:

kinetic energy =potential energy

½ (m+M)V = (m+M)gh.

So, the speed of the system immediately after the collision is:

V = (2gh)½

0.28=(2*9.81*h)^0.5

0.0784=2*9.81h

h=0.00399m

8 0

4 years ago

An object, initially at rest moves 250m in 17s. What is it's acceleration?

Gekata [30.6K]

With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity

d= 250s
t= 17s

a= d/t
= $\frac{250}{17}$
= 4.7 $\frac{m}{s}$

8 0

3 years ago

Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note

love history [14]

Answer:

The energy is $9.4\times10^{-21}\ J$

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

$Q=\Delta U+W$

$Q=\Delta U+nRT$

Put the value into the formula

$4480=\Delta U+\dfrac{20}{28}\times8.314\times77$

$\Delta U=4480-\dfrac{20}{28}\times8.314\times77$

$\Delta U=4022.73\ J$

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

$N=n\times \text{Avogadro number}$

Put the value into the formula

$N=\dfrac{20}{28}\times6.02\times10^{23}$

$N=4.3\times10^{23}$

We need to calculate the energy

Using formula of energy

$E=\dfrac{\Delta U}{N}$

Put the value into the formula

$E=\dfrac{4022.73}{4.3\times10^{23}}$

$E=9.4\times10^{-21}\ J$

Hence, The energy is $9.4\times10^{-21}\ J$

4 0

3 years ago

It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibilty can be described by Hooke

Varvara68 [4.7K]

<span>k = 1.7 x 10^5 kg/s^2 Player mass = 69 kg Hooke's law states F = kX where F = Force k = spring constant X = deflection So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration. F = kX F/X = k 115 kg* 9.8 m/s^2 / 0.65 cm = 115 kg* 9.8 m/s^2 / 0.0065 m = 1127 kg*m/s^2 / 0.0065 m = 173384.6154 kg/s^2 Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2 Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So X/0.39 cm = 115 kg/0.65 cm X = 44.85 kg/0.65 X = 69 kg The player masses 69 kg.</span>

4 0

4 years ago

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