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podryga [215]
2 years ago
11

Consider the mass and velocity values of Objects A and B below.

Physics
1 answer:
777dan777 [17]2 years ago
8 0

Answer:

the same

Explanation:

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Please help.....Which of the following is most likely the distance between the Milky Way and another galaxy?
bagirrra123 [75]

Answer:

The answer would be 10.8 billion light years

6 0
3 years ago
A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs
defon

Answer:

2940.1 joules  would you burn in climbing stairs all day.

Explanation:

Work = W = F\times d

going up stairs  would be against force of gravity

W = mgh

where h is the  height

the question is not complete because we need speed or distance

h =  v \times t

so assuming 1 step per second

h = 86,400 steps  \times 7inchs/step \times 0.0254 m/inch

h = 15362 m

so from this    

W = 800 N \times 15362

   = 12289600 J  

that means YOU  need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal  

Kcal =  \frac{W}{(J/Kcal)}

       = \frac{12289600}{4180}

       = 2940.1 Kcal

if you ran faster you would use more energy  2 steps per second would mean 5880 Kcal.

8 0
3 years ago
A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const
grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr.  and Area of this element is

dA=2\pi r dr

since current density is given

J=kr

then , current through this element will be,

di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr

integrating on both sides between the appropriate limits,

\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr&#10;\\\\&#10;I=\frac{2\pi\,ka^3}{3} -------------------------------(1)

Magnetic field can be found by using Ampere's law

\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=&#10;\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}&#10;\\\\B_{in}=\frac{\mu_0kl^2}{3}&#10;

now using equation 1, putting the value of k,

B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}&#10;\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}&#10;

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)&#10;\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}&#10;\\\\B_{out}=\frac{\mu_0ka^3}{3r}&#10;

again using,equaiton 1,

B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}&#10;\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}

8 0
3 years ago
A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of
Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =  <u>      mass    </u>

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0
3 years ago
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Sowers has spent her career fighting for gender equality and it passionate about it .
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