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Anastaziya [24]
3 years ago
10

A ball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of

time it takes to return to earth
Physics
1 answer:
Dafna11 [192]3 years ago
7 0
In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)

d = 8,848m and g = 9.8m/s

Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)

The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)

Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)

Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

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intial velocity is 45km/h, final velocity is 0, time taken is 3 seconds,calculate the retradation of the vehicle and distance tr
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A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current
Tema [17]

Answer:410.90\times 10^{-6} V

Explanation:

Given

A=5.5 mm^2

n=919 turns/cm\approx 85400 turns/m

\omega =226 rad/s

According to the Faraday law of induction, induced emf is given by

E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}

Magnetic field \phi _B=B\cdot A

B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )

B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)

B=0.3305\sin (226t)

\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}

\phi _{B}=1.818\times 10^{-6}\sin (226t)

E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}

E=-1.818\times 10^{-6}\times 226\cos (226t)

E=-410.90\times 10^{-6}\cos (226t)

Amplitude of EMF induced=410.90\times 10^{-6} V

8 0
3 years ago
A technician wearing a brass bracelet enclosing area 0.00500 m2 places her hand in a solenoid whose magnetic field is 3.10 T dir
aliya0001 [1]

Explanation:

Given that,

Area enclosed by a brass bracelet, A=0.005\ m^2

Initial magnetic field, B_i=3.1\ T

The electrical resistance around the circumference of the bracelet is, R = 0.02 ohms

Final magnetic field, B_f=0.93\ T

Time, t=16\ ms=16\times 10^{-3}\ s

The expression for the induced emf is given by :

\epsilon=-\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_f-B_i}{t}

\epsilon=-0.005 \times \dfrac{0.93-3.1}{16\times 10^{-3}}

\epsilon=0.678\ volts

So, the induced emf in the bracelet is 0.678 volts.

Using ohm's law to find the induced current as :

V = IR

I=\dfrac{V}{R}

I=\dfrac{0.678}{0.02}

I = 33.9 A

or

I = 34 A

So, the induced current in the bracelet is 34 A. Hence, this is the required solution.

5 0
3 years ago
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