Answer:
2 per s
Explanation:
divid 40 and 20 it gives you = 2
The centripetal acceleration of an object is given by the relation,
![Ac =V^2/R](https://tex.z-dn.net/?f=Ac%20%3DV%5E2%2FR)
where Ac = centripetal acceleration =
R = radius of rotation = 15 m
V = speed of astronaut
Hence, ![\frac{V^2}{15} =98](https://tex.z-dn.net/?f=%5Cfrac%7BV%5E2%7D%7B15%7D%20%3D98)
solving this we get, V = 38.34 m/s
Answer:
![\large \boxed{\text{761 kJ}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B761%20kJ%7D%7D)
Explanation:
You calculate the energy required to break all the bonds in the reactants.
Then you subtract the energy needed to break all the bonds in the products.
N₂ + O₂ ⟶ 2NO
N≡N + O=O ⟶ 2O-N=O
Bonds: 2N≡N 1O=O 2N-O + 2N=O
D/kJ·mol⁻¹: 941 495 201 607
![\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta%20H%20%26%20%3D%20%26%20%5Csum%7BD_%7B%5Ctext%7Breactants%7D%7D%7D%20-%20%5Csum%7BD_%7B%5Ctext%7Bproducts%7D%7D%7D%5C%5C%26%20%3D%20%26%202%20%5Ctimes%20941%20%2B1%20%5Ctimes%20495%20-%20%282%20%5Ctimes%20201%20%2B%202%5Ctimes%20607%29%5C%5C%26%3D%26%202377%20-%201616%5C%5C%26%3D%26%5Ctextbf%7B761%20kJ%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20enthalpy%20of%20reaction%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B761%20kJ%7D%7D%24%7D.)
Distance and time, distance because that's how far and time because that's how long
Answer
given,
Speed of car A = 95 Km/h
= 95 x 0.278 = 26.41 m/s
Speed of Car B = 121 Km/h
= 121 x 0.278 = 33.64 m/s
Distance between Car A and B at t=0 = 41 Km
a) Distance travel by car B
d = 26.41 t + 41000
speed of the car A = 33.64 m/s
distance = s x t
26.41 t + 41000 = 33.64 x t
7.23 t = 41000
t = 5670.82 s
time taken by Car B to cross Car A is equal to t = 5670.82 s
distance traveled by car A
D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km
b) distance travel by the car B in 30 s after overtaking car A
D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km