An increase in average speed of 1 km/h typi- cally results in a 3% higher risk of a crash involving injury, with a 4–5% increase for crashes that result in fatalities. — Speed also contributes to the severity of the impact when a collision does occur. Hope this helps!
My Very Excellent Mother Just Served Us Nachos
(or Nine pickles if Pluto was a planet)
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
(Pluto)
hope that helped
The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
Therefore, option A is correct option.
Given,
Mass m = 14g
Volume= 3.5L
Temperature T= 75+273 = 348 K
Molar mass of CO = 28g/mol
Universal gas constant R= 0.082057L
Number of moles in 14 g of CO is
n= mass/ molar mass
= 14/28
= 0.5 mol
As we know that
PV= nRT
P × 3.5 = 0.5 × 0.082057 × 348
P × 3.5 = 14.277
P = 14.277/3.5
P = 4.0794 atm
P = 4.1 atm.
Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
learn more about pressure:
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Well based on the three I would think it's C
Answer:
(a) 1.73 s
(b) 14.75 m
(c) 3.36 s
(d) double
(e) 63.32 m
Explanation:
Vertical component of initial velocity, uy = 17 m/s
Horizontal component of initial velocity, ux = 18.3 m/s
(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.
Use first equation of motion in vertical direction
vy = uy - gt
0 = 17 - 9.8 t
t = 1.73 seconds
(B) Let the highest point is at height h.
Use III equation of motion in vertical direction

0 = 17 x 17 - 2 x 9.8 x h
h = 14.75 m
(C) The time taken by the ball to return to original level is T.
Use second equation of motion i vertical direction.

h = 0 , u = 17 m/s
0 = 17 t - 0.5 x 9.8 t^2
t = 3.46 second
(D) It is the double of time calculated in part A
(E) Horizontal distance = horizontal velocity x total time
d = 18.3 x 3.46 = 63.32 m