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elena-s [515]
3 years ago
8

Plants take up carbon dioxide from the air and nutrients from the soil. This is an example of interactions between the

Physics
1 answer:
Vitek1552 [10]3 years ago
4 0
Air and the chemicals in the oxgen
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Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl
ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
  • Both movements are independent each other, due to they are perpendicular.
  • In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
  • Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

       v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)

  • Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s  (2)

  • In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
  • Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       \Delta y = v_{oy}  * t - \frac{1}{2} *g*t^{2} (3)

  • In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)

  • Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       \Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)

  • Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
4 0
3 years ago
An unusually high tide is called a _______ tide.
Monica [59]
 Proxigean Spring <span>Tide</span>
8 0
3 years ago
A syringe of volume 16 cm3 is filled with air to a pressure of 1.03 atm. If the piston of the syringe is pushed to change the vo
denpristay [2]
<h3>Answer:</h3>

189.07 kPa

<h3>Explanation:</h3>

Concept tested: Boyle's law

<u>We are given;</u>

  • Initial volume of the syringe, V1 is 16 cm³
  • Initial pressure of the syringe, P1 is 1.03 atm
  • New volume of the syringe, V2 is 8.83 cm³

We are required to calculate the new pressure of the syringe;

  • We are going to use the concept on Boyle's law of gases.
  • According to the Boyle's law, for a fixed mass of a gas, the pressure is inversely proportional to its volume at constant temperature.
  • That is; P α 1/V
  • At varying pressure and volume, k(constant)  = PV and P1V1=P2V2

Therefore, to get the new pressure, P2, we rearrange the formula;

P2 = P1V1 ÷ V2

     = ( 16 cm³ × 1.03 atm) ÷ 8.83 cm³

    = 1.866 atm.

  • Thus, the new pressure is 1.866 atm
  • But, we need to convert pressure to Kpa
  • Conversion factor is 101.325 kPa/atm

Thus;

Pressure = 1.866 atm × 101.325 kPa/atm

               = 189.07 kPa

Hence, the new pressure of the air in the syringe is 189.07 kPa

3 0
3 years ago
An object is observed for a time interval of 20 seconds. From time 6.7 s the object experiences a Force of 106 N that lasts unti
shusha [124]

Answer:

1654 kg m/s

Explanation:

The impulse experienced by an object is equal to the product between the force exerted on the object and the time during which the force lasts:

I=F\Delta t

where:

I is the impulse

F is the force exerted on the object

\Delta t is the time during which the force is applied

For the object in this problem, we have

F=106 N (force applied)

\Delta t= 15.6 s (time interval)

Therefore, the impulse experienced by the object is:

I=(106)(15.6)=1654 kg m/s

3 0
3 years ago
When you polish a window with a dry cloth on a dry day it becomes dusty why does this not happen on a wet day
barxatty [35]

When glass is rubbed with a dry cloth, the friction creates charged static electricity; this in turn attracts small non charged particles of dust. The simplest way to put it, the dry cloth creates a static charge that attracts non charged dust particles.

3 0
3 years ago
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