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3 years ago

A 75 kg pilot flies a plane in a loop. At the top of the loop, where the plane is completely

Physics

1 answer:

erica [24]3 years ago

6 0

Answer:

1.47 km

Explanation:

When the pilot hangs freely in the seat and does not push against the seat belt, his centripetal acceleration would counter balance gravitational acceleration g = 9.81m/s2 = a

We can calculate the radius of the loop using velocity v = 120m/s and a

$a = \frac{v^2}{r}$

$r = \frac{v^2}{a} = \frac{120^2}{9.81} = 1467.89 m$ or 1.46 km

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Rank the following in order of increased size? A. Proton B. Nucleus C. Electron D. Atom

Basile [38]

From largest to smallest- atom, nucleus, proton and electron

7 0

2 years ago

A ship sets out to sail to a point 154 km due north. An unexpected storm blows the ship to a point 72 km due east of its startin

Norma-Jean [14]

Answer:

Explanation:

If a ship sets out to sail to a point 154 km due north and an unexpected storm blows the ship to a point 72 km due east of its starting point, then the ships distance from the original destination can be gotten by finding the displacement of the ship and this can be gotten by using pythagoras theorem.

Let D be the unknown displacement

According to the theorem;

D² = 154² + 72²

D² = 23716 + 5184

D² = 28900

D = √28900

D = 170km

<em>This means that the ship must now sail a distance of 170km for it to reach its original destination.</em>

8 0

3 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. You have a stopped pipe of adjustable length clo

faltersainse [42]

Answer:

Length of pipe $= 0.057$ meter

Explanation:

Speed of a transverse wave on a string

$v = \sqrt{\frac{F}{\mu} }$

where F is the tension in string and $\mu$ is the mass per unit length

Thus,

$\mu = \frac{m}{L}$

Substituting the given values we get -

$\mu = \frac{7.25 * 10^{-3}}{0.62}\\mu = 0.0117 \frac{Kg}{m}$

Speed of a transverse wave on a string

$v = \sqrt{\frac{4510}{0.0117} } \\v = 620.86 \frac{m}{s}$

For third harmonic wave , frequency is equal to

$f = \frac{nv}{2L}$

Substituting the given values, we get -

$f = \frac{3 * 620.86}{2 * 0.62} \\f = 1502.08$

Length of pipe

$L = \frac{nv}{4 f}$

Substituting the given values we get

$n = 1$ for first harmonic wave

$L = \frac{344* 1}{4*1502.08} \\L = 0.057$

Length of pipe $= 0.057$ meter

5 0

3 years ago

A car initially at rest undergoes uniform acceleration for 6.32 seconds and covers a distance of 120 meters. What is the approxi

Verizon [17]

Using kinematic equation s=ut + 1/2 at^2(u = initial velocity=0, s=120m, t= 6.32s), 120 = 0(t) + 1/2 a(6.32)^2. a = 120x2/(6.32)^2 = 6m/s^2.

3 0

3 years ago

Read 2 more answers

a single 1,300 jk cargo car is rolling along a train track at 2.0 m/s when 400 kg of coal is dropped vertically into it. what is

tensa zangetsu [6.8K]

Answer:

1.53 m/s

Explanation:

Given:

Mass of the car (M) = 1300 kg

Mass of the coal (m) = 400 kg

Initial velocity of the car (U) = 2 m/s

Initial velocity of the coal (u) = 0 m/s (Since it is dropped)

When the coal is dropped into the car, then they move with same final velocity.

Let the final velocity be 'v' m/s.

For a closed system, the law of conservation of momentum holds true.

So, initial momentum is equal to final momentum of the car-coal system.

Initial momentum of the car = $MU=1300\times 2=2600\ Ns$

Initial momentum of the coal = $mu=0\ Ns$

Total initial momentum is the sum of the above two momentums.

So, total initial momentum = 2600 + 0 = 2600 Ns

Now, final momentum is given as the product of combined mass and final velocity. So,

Final momentum of the system = $(M+m)v=(1300+400)v=1700v$

Now, from law of conservation of momentum,

Initial momentum = Final momentum

$2600=1700v\\\\v=\frac{2600}{1700}\\\\v=1.53\ m/s$

Therefore, the final velocity of either of the two masses is same is equal to 1.53 m/s.

4 0

3 years ago