Let suppose that the jet liner accelerates uniformly. From statement we know the initial ( $v_{o}$ ) and final speeds ( $v_{f}$ ), measured in meters per second, of the aircraft and likewise the runway length ( $d$ ), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed ( $a$ ), measured in meters per square second:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 82\,\frac{m}{s}$ and $d = 1500\,m$ , then the acceleration that the jet must have is:

$a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}$

$a = 2.241\,\frac{m}{s^{2}}$

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0

2 years ago

Explain resolution of Force

Margarita [4]

Answer:

it is defined as splitting up the given force into a number of components, without changing its effects on the body is called resolution of forces. A force is generally resolved along with two mutually perpendicular directions.

Explanation:

7 0

2 years ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

4 0

3 years ago