A decrease in the magnitude of velocity is called

You are in Paris, 60 m up in the Eiffel Tower. If you throw a euro downward at a velocity of 2.0 m/s, how long would it take the

kondor19780726 [428]

Answer:

t = 3.29 seconds

Explanation:

It is given that,

Height of the Eiffel tower is 60 m

Initial speed of a euro, u = 2 m/s

It will move under the action of gravity in the downward direction. Firstly, we can find the final velocity as follows :

$v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad} \\\\v=\sqrt{(2)^2+2\times 9.81\times 60} \\\\v=34.36\ m/s$

Let t is the time taken by the euro to hit the ground. It can be calculated as :

$v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{34.36-2}{9.81}\\\\t=3.29\ s$

Hence, it will take 3.29 seconds to hit the ground.

4 0

3 years ago

D Question 14 2 pts The universe could be considered an isolated system because

Blizzard [7]

Answer:

it would help if we knew the question and other answers

Explanation:

5 0

3 years ago

A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).

yawa3891 [41]

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

$f = n\frac{v}{2L}$

Where

v = speed of sound

L = Length of the pipe

n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

Re-arrange to find L,

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

The radius between the two frequencies would be 4 to 5,

$\frac{528Hz}{660Hz}= \frac{4}{5}$

$4:5$

Therefore the frequencies are in the ratio of natural numbers. That is

$4f = 528\\f = \frac{528}{4}\\f = 132Hz$

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

$L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m$

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

$L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m$

Therefore the length of the pipe in the second harmonic is 2.6m

7 0

3 years ago

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav

Dmitrij [34]

Answer:

$t=1.5\times 10^{-4}\ s$

Explanation:

Given:

file size to be transmitted, $D=30\ Mb$

transmission rate of data, $\dot D=10\ Mb.s^{-1}$

propagation speed, $v=2\times 10^8\ m.s^{-1}$

distance of data transfer, $s=10000\ km=10^4\ m$

Now the delay in data transfer from source to destination for each 10 Mb:

$t'=\frac{s}{v}$

$t'=\frac{10^4}{2\times 10^8}$

$t'=5\times 10^{-5}\ s$

Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:

So,

$t=3\times t'$

$t=3\times 5\times 10^{-5}$

$t=1.5\times 10^{-4}\ s$

3 0

3 years ago

These show how something works that can't be seen

Wewaii [24]

The answer is cells.

6 0

3 years ago