A planetary nebula is the phase of a

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

A 2 kg object being pulled across the floor with a speed of 10 m/sec is suddenly

zvonat [6]

Answer:

The frictional force producing this deceleration would have a magnitude of $4\; \rm N$ .

Explanation:

The velocity of this object changed by $\Delta v = (-10\; \rm m\cdot s^{-1})$ in $\Delta t = 5\; \rm s$ . The acceleration of this object would be:

$\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}$ .

Let $m$ denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:

$\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}$ .

( $1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}$ .)

If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be $4\; {\rm N}$ , same as the magnitude of the net force on this object.

5 0

3 years ago

_mg(oh)2 + _HBr _mgBr2 + _HOH

s344n2d4d5 [400]

A) that's not physics, that's chemistry B) assuming you want it balanced, 1,2,1,2

3 0

3 years ago

A train starts from a station with a constant acceleration of at = 0.40 m/s2. A passenger arrives at the track time t = 6.0s aft

jok3333 [9.3K]

Answer:

4.8 m/s

Explanation:

When she catches the train,

They will have travelled the same distance.and
Their speeds will be equal

The formula for the distance covered by the train is

d = ½at² = ½ × 0.40t² = 0.20t²

The passenger starts running at a constant speed 6 s later, so her formula is

d = v(t - 6.0)

The passenger and the train will have covered the same distance when she has caught it, so

(1) 0.20t² = v(t - 6.0)

The speed of the train is

v = at = 0.40t

The speed of the passenger is v.

(2) 0.40t = v

Substitute (2) into (1)

0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t

Subtract 0.20t² from each side

0.20t² - 2.4t = 0

Factor the quadratic

t(0.20t - 2.4) = 0

Apply the zero-product rule

t =0 0.20t - 2.4 = 0

0.20t = 2.4

(3) t = 12

We reject t = 0 s.

Substitute (3) into (2)

0.40 × 12 = v

v = 4.8 m/s

The slowest constant speed at which she can run and catch the train is 4.8 m/s.

A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.