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Arlecino [84]
3 years ago
12

A rocket blasts off. In 10.0 seconds it is at 10,000 ft, traveling at 3600 mph. Assuming the direction is up, calculate the acce

leration. (Hint: the rocket is not under constant acceleration). 5280 ft/s2 528 ft/s2 100 ft/s2 200. ft/s2
Physics
2 answers:
marissa [1.9K]3 years ago
7 0

       Acceleration  =  (change in speed)  /  (time for the change)

Change in speed = (ending speed) - (beginning speed)

                            =     (3,600 mi/hr) - ( 0 )

                            =       3,600 mi/hr

                            =     same as  1 mile/second.

Acceleration  =  (1 mi/sec) / (10 sec)

                      =     0.1 mi/sec² .

But 1 mile = 5,280 ft,
so
                       0.1 mi/s²  =  528 ft/s²          
mojhsa [17]3 years ago
7 0

Answer:

 0.1 mi/s²  =  528 ft/s²    

Explanation:

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3 years ago
When we breathe, we inhale oxygen and exhale carbon dioxide plus water vapor. Which likely has more mass, the air that we inhale
Solnce55 [7]

Answer:

If the same volume of air is inhaled and exhaled, the air we breathe out normally weighs more than the air we breathe in.

Since the output from the body normally exceeds the input, breathing leads to weight loss.

Explanation:

If equal volumes of gas is inhaled and exhaled, the exhaled gas is heavier.

The inhaled gas contains Oxygen and majorly Nitrogen.

The exhaled gas contains CO₂, H₂O and a very large fraction of the unused inhaled air that goes into the lungs.

So, basically, the body exchanges O₂ with CO₂ and H₂O (and some other unwanted gases in the body) in a composition that CO₂, the heavier gas of the ones mentioned here, is prominent.

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5 0
3 years ago
A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,00
Savatey [412]

Answer:

The answer is "Option B".

Explanation:

r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\

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7 0
3 years ago
A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 g
alekssr [168]

Answer:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                  y cm  

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

              12cm             50 cm              69cm

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                 19 cm                                              

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   

5 0
3 years ago
Which type of wave would not be classified as a transverse wave?
Jlenok [28]
Longitud wave something like that.
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4 years ago
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