Ask question

Ask question

All categories

3 years ago

14

Neglecting air resistance, if a package is dropped from an airplane flying 100 km/hr at an altitude of 200 meters, how far from

the drop point would the package land? Give answer in meters.

1 answer:

UkoKoshka [18]3 years ago

4 0

Answer:

178 m

Explanation:

You might be interested in

Witch part of the ear is responsible for sending singnals about sound to the brain

drek231 [11]

Answer:

Cochlea

Explanation:

Cochlea. The $cochlea$ changes the sound vibrations from the middle ear into nerve signals. The nerve signals the travel to the brain via the $cochlea \ nerve$ .

The cochlea nerve is also called the auditory nerve.

7 0

2 years ago

A long, hollow wire with inner radius a and outer radius b carries a uniform current density J. What is the magnetic field as a

sveta [45]

Answer:

The magnetic field in the region a < r < b is $B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

Explanation:

If we have the a < r < b. The formula of current is:

$J=\frac{I_{total} }{A}$

Where:

A = area enclosed by the loop.

Itotal = total current in loop.

$J=\frac{I}{\pi b^{2}-\pi a^{2} }$

$I_{enclosed} =JA_{enclosed}$

$I_{enclosed} =\frac{I(\pi r^{2}- \pi a^{2})}{\pi b^{2}-\pi a^{2} }$

If we have the Ampere`s law:

$\int\limits^a_b {B} \, ds =u_{0} I_{enclosed} \\2B\pi r=u_{0} (\frac{I(\pi r^{2}-\pi ^{2} }{\pi ^{2}-\pi a^{2} } )\\B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

6 0

2 years ago

Local environmental change have no effect on the global climate true or false

Otrada [13]

Answer:

false

Explanation:

Even small environmental changes have large effects because small things become bigger

5 0

2 years ago

Dominik [7]

Answer:

i Believe the correct answer is "An open circuit being changed into a closed circuit"

Explanation:

8 0

2 years ago

Two violinists are trying to tune their instruments in an orchestra. One is producing the desired frequency of 440.0 hz. The oth

Katyanochek1 [597]

Answer:

Percentage change in tension is 3.8%

Explanation:

We have given initially frequency $f_1$ = 440 Hz

Let tension in the string at this frequency is $T_1$

Now second frequency is $f_2=448.4Hz$

Frequency in string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

From the relation we can say that

$\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}$

$\frac{440}{448.4}=\sqrt{\frac{T_1}{T_2}}$

${\frac{T_2}{T_1}}=1.038$

Percentage change in tension is equal to

$=\frac{T_2-T_1}{T_1}=\frac{T_2}{T_1}-1=(1.038-1)\times 100=3.8$ %

So percentage change in tension is 3.8%

4 0

2 years ago