Ask question

Ask question

All categories

bagirrra123 [75]

1 year ago

9

The eiffel tower is 300 meters tall. Disregarding air friction, at what velocity would an object be traveling when it reaches th

e ground if it were dropped from the top of the tower?.

1 answer:

xeze [42]1 year ago

8 0

Answer:

Hope it helps you :)

Explanation in the pic above.

You might be interested in

Which is the best example of a cycle: a turn of a wheel or a slide down a ski slope?

melisa1 [442]

Definitiely a turn of a wheel. Skiing down a ski slope is going down at an angle and it's usually straight. The turn of a wheel just makes a lot more sense

5 0

3 years ago

g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

$\frac{R}{2L}T= \ln \frac{5}{3.8}$

$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

= 11.45

4 0

2 years ago

Hey if your answering this can you just answer the number 14 I already got the rest<br> thanks

Scrat [10]

Layer a is the answer

7 0

2 years ago

Read 2 more answers

Determine the index of refraction of glass that is struck by unpolarized light at 53.8 degrees and resulting in a fully polarize

nignag [31]

Answer:

The refractive index of glass, $\mu_{g} = 1.367$

Solution:

Brewster angle is the special case of incident angle that causes the reflected and refracted rays to be perpendicular to each other or that angle of incident which causes the complete polarization of the reflected ray.

To determine the refractive index of glass:

$tan\theta_{P} = \frac{\mu_{g}}{\mu_{a}}$ (1)

where

$\mu_{a}$ = refractive index of glass

$\mu_{g}$ = refractive index of glass

Now, using eqn (1)

$tan{53.8^{\circ}} = \frac{\mu_{g}}{1}$

$\mu_{g} = tan53.8^{\circ}$

$\mu_{g} = 1.367$

5 0

2 years ago

A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a

Burka [1]

Answer:

350 N/m

Explanation:

If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m

k = 70 / 0.2 = 350 N/m

The spring constant is 350 N/m

4 0

3 years ago

Read 2 more answers