Answer:
False
Explanation:
False
If brakes fail then we should not run against any object or wall to slow down the speed instead we should use friction forces slow down the car and shift gear to neutral to cease any acceleration.
By driving zig-zag motion friction force on the tires may stop the car after some distance.
Do not try to shut off the engine as it will lock the steering mechanism any prove fatal.
An airplane wing is designed to make the air move <em>faster over the top than under the bottom. (c)</em> They do this by making the top of the wing curved and the bottom flat, so the air that flows over the top has farther to go.
It turns out, according to a scientist named Bernoulli, that air exerts lower pressure when it moves faster. So if it moves faster over the top of the wing, then there's less pressure on the top of the wing and more on the bottom, and the wing thinks there's a force that's LIFTING it . . . <em>very convenient and useful</em> if the wing is part of an airplane !
Answer:
Part A:
to two significant figures
Part B:
to two significant figures
Part C:
to two significant figures
Explanation:
Given that :
mass of the hydrogen = 0.30 g
the molar mass of hydrogen gas molecule = 2 g/mol
we all know that:
number of moles = mass/molar mass
number of moles = 0.30 g /2 g/mol
number of moles = 0.15 mol
For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = ![C_v=\dfrac{3}{2}R](https://tex.z-dn.net/?f=C_v%3D%5Cdfrac%7B3%7D%7B2%7DR)
For Part A:
![Q = mC_v\Delta T](https://tex.z-dn.net/?f=Q%20%3D%20mC_v%5CDelta%20T)
![Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K](https://tex.z-dn.net/?f=Q%3D%200.15%20%5C%20mol%20%28%5Cdfrac%7B3%7D%7B2%7D%29%288.314%20%5C%20J%2Fmole.K%20%29%28100-50%29K)
![Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)](https://tex.z-dn.net/?f=Q%3D%200.15%20%5Ctimes%20%28%5Cdfrac%7B3%7D%7B2%7D%29%20%5Ctimes%20%288.314%20%5C%20J%20%29%5Ctimes%20%2850%29)
![Q=93.5325 \ J](https://tex.z-dn.net/?f=Q%3D93.5325%20%5C%20J)
to two significant figures
Part B. For hot temperature, ![C_v=\dfrac{5}{2}R](https://tex.z-dn.net/?f=C_v%3D%5Cdfrac%7B5%7D%7B2%7DR)
![Q = mC_v\Delta T](https://tex.z-dn.net/?f=Q%20%3D%20mC_v%5CDelta%20T)
![Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K](https://tex.z-dn.net/?f=Q%3D%200.15%20%5C%20mol%20%28%5Cdfrac%7B5%7D%7B2%7D%29%288.314%20%5C%20J%2Fmole.K%20%29%28300-250%29K)
![Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)](https://tex.z-dn.net/?f=Q%3D%200.15%20%5Ctimes%20%28%5Cdfrac%7B5%7D%7B2%7D%29%20%5Ctimes%20%288.314%20%5C%20J%20%29%5Ctimes%20%2850%29)
![Q=155.8875 \ J](https://tex.z-dn.net/?f=Q%3D155.8875%20%5C%20J)
to two significant figures
Part C. For an extremely hot temperature, ![C_v=\dfrac{7}{2}R](https://tex.z-dn.net/?f=C_v%3D%5Cdfrac%7B7%7D%7B2%7DR)
![Q = mC_v\Delta T](https://tex.z-dn.net/?f=Q%20%3D%20mC_v%5CDelta%20T)
![Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K](https://tex.z-dn.net/?f=Q%3D%200.15%20%5C%20mol%20%28%5Cdfrac%7B7%7D%7B2%7D%29%288.314%20%5C%20J%2Fmole.K%20%29%282300-2250%29K)
![Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)](https://tex.z-dn.net/?f=Q%3D%200.15%20%5Ctimes%20%28%5Cdfrac%7B7%7D%7B2%7D%29%20%5Ctimes%20%288.314%20%5C%20J%20%29%5Ctimes%20%2850%29)
![Q=218.2425 \ J](https://tex.z-dn.net/?f=Q%3D218.2425%20%5C%20J)
to two significant figures
Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>
Answer:
The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.
Explanation:
Let us call the mass of the balloon
and the mass of the basket
, then according to newton's second law:
,
where
is the upward acceleration, and
is the net propelling force (counts the gravitational force).
Also, the tension
in the rope is 79.8 N more than the basket's weight; therefore,
![(2). \:T = m_2g+79.8](https://tex.z-dn.net/?f=%282%29.%20%5C%3AT%20%3D%20m_2g%2B79.8)
and this tension must equal
![T -m_2g =m_2a](https://tex.z-dn.net/?f=T%20-m_2g%20%3Dm_2a)
![(3). \:T = m_2g +m_2a](https://tex.z-dn.net/?f=%283%29.%20%5C%3AT%20%3D%20m_2g%20%2Bm_2a)
Combining equations (2) and (3) we get:
![m_2a = 79.8](https://tex.z-dn.net/?f=m_2a%20%3D%2079.8)
since
, we have
![\boxed{m_2 = 301.13kg}](https://tex.z-dn.net/?f=%5Cboxed%7Bm_2%20%3D%20301.13kg%7D)
Putting this into equation (1) and substituting the numerical values of
and
, we get:
![688N = (m_1+301.13kg)(0.265m/s^2)](https://tex.z-dn.net/?f=688N%20%3D%20%28m_1%2B301.13kg%29%280.265m%2Fs%5E2%29)
![\boxed{m_1 = 2295 kg}](https://tex.z-dn.net/?f=%5Cboxed%7Bm_1%20%3D%202295%20kg%7D)
Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.
Using current technology, useful parallax measurements can only be found for stars up to about 340 light years (100 parsecs) away.