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3 years ago

POSSIE

Physics

1 answer:

kotegsom [21]3 years ago

3 0

Answer:

please find attached pdf

Explanation:

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An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. the acceleration

Yanka [14]

The acceleration due to gravity serves as the centripetal acceleration of the objects that orbits the Earth. The centripetal acceleration due to gravity is calculated through the equation,

a = v²/r

where v is the speed and r is the radius. Substituting the known values to the equation,

9.8 m/s² = (420 m/s)² / r

The value of r from the equation is 18000 m or equal to 18 km.

<em>Answer: 18 km</em>

6 0

3 years ago

A typical wall outlet voltage in the United States is 120 volts. Personal MP3 players require much smaller voltages, typically 4

alexgriva [62]

Answer:

Number of turns on the secondary coil of the adapter transformer is 10.

Explanation:

For a transformer,

$\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }$

where $V_{s}$ is the voltage induced in the secondary coil

$V_{p}$ is the voltage in the primary coil

$N_{s}$ is the number of turns of secondary coil

$N_{p}$ is the number of turns of primary coil

From the given question,

$\frac{487*10^{-3} }{120}$ = $\frac{N_{s} }{2464}$

⇒ $N_{s}$ = $\frac{2462*487*10^{-3} }{120}$

= 9.999733

∴ $N_{s}$ = 10 turns

5 0

3 years ago

Read 2 more answers

What type of collision is being described in each statement.

Ostrovityanka [42]

Answer:

inelastic, since the girl moves in the same direction as the thrown ball

Explanation:

yess this ok

UwU

7 0

2 years ago

Find the kinetic energy of a 0.1 kg toy truck moving at a speed of 1.1 m/s

enot [183]

0.0605J is your answer. Use the formula KE=1/2mv^2

8 0

3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$