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GaryK [48]
3 years ago
8

How are airplanes artificially pressurized?

Physics
1 answer:
Irina18 [472]3 years ago
8 0
They are built to be airtight so thsy sir can't get in or out, and then, as they climb up to where the air is thin, they use an air pump to pimp air into the inside.

Just like a party balloon.
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PLEASE HELP!!
leva [86]
I think the answer is B
8 0
2 years ago
tickets at a museum cost 17 dollars each for a field trip, the museum offers a 4 doller discount on each ticket. How much will t
Marizza181 [45]

Answer:

$416 sounds like the best answer.

Explanation:

The tickets started off at $17, but a $4 discount for EACH ticket bought.

so 17-4 is 13

now tickets cost $13 each.

Multiply it by 32.

13x32=416

5 0
3 years ago
A vector must always have both size and ..
DochEvi [55]
Both magnitude and DIRECTION
For example,
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5 0
3 years ago
I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite
e-lub [12.9K]

Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as

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

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

substituting values

                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

3 0
3 years ago
What best describes gravity?
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 F = G*((m sub 1*m sub 2)/r^2) 
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