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Zanzabum
3 years ago
10

Assume that the equator of the Earth is 24,000 miles long (that’s its circumference). Now, pretend that you are standing somewhe

re on the equator, such as in the country of Ecuador. Now, if the Earth turns once, completely, in 24 hours, then, how fast would you be going (in miles per hour), even if you just stood still?
Physics
1 answer:
slega [8]3 years ago
3 0

Answer:

1000 mph

Explanation:

P = Perimeter = 24000 mi

r = Radius of the equator

t = Time taken to complete one rotation = 24 h

Perimeter of a circle is given by

P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi

Angular speed is given by

\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}

Velocity if given by

v=r\omega\\\Rightarrow v=\dfrac{24000}{2\pi}\times \dfrac{2\pi}{24}\\\Rightarrow v=1000\ mph

The person would be going at a speed of 1000 mph

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lara31 [8.8K]

Answer:

The values is  B  = 3.2 *10^{-8} \  T

The  direction is out of the plane

Explanation:

From the question we are told that

  The  magnitude of the electric field is  E  =  9.6 \  V/m

 

The  magnitude of the magnetic field is mathematically represented as

       B  = \frac{E}{c}

where c is the speed of light with value

      B  = \frac{ 9.6}{3.0 *10^{8}}

     B  = 3.2 *10^{-8} \  T

Given that the direction off the electromagnetic wave( c ) is  northward(y-plane ) and  the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page  (z-plane  )

     

3 0
3 years ago
THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

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2 years ago
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Answer:

Explanation:

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Where;

m = mass (kg)

g = acceleration due to gravity (10m/s²)

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