Adragster with a mass of 700 kg attains a speed of 120 m/s in the quarter mile. immediately after passing the timing lights, the driver opens the drag chute, of area 7.5 m2. air and rolling resistance of the car may be neglected. find the time required for the machine to reduce its velocity to 20 m/s in standard air.

The time taken is 5.55 seconds

Explanation:

Density, p = 1kg/m³

Mass of vehicle, m = 700kg

Initial speed, u = 120m/s

Area of drag chute, A = 7.5 m²

Drag coefficient, Cd = 1.4

Final velocity, v = 20m/s

time taken to decelerate to 20m/s, t = ?

EF = ma

Since the force taken into consideration is the drag force due to the drag chute,

EF = -Fd = -(pACdv²)/2 = ma

a = dv/dt

So the equation can be written as,

-(pACdv²)/2 = m dv/dt

Taking integral of the left hand side with respect to dt and the right with respect to dv with boundaries of 0 to t and u to v,

(pACd)t/2m = (1/v) - (1/u)

t = (2m/pACd) x ( (1/v) - (1/u) ) = f(V)

At V = v

t = (2 x 700)/(1)(7.5)(1.4) x (1/20) - (1/120)

t = 5.55 seconds

number of electrons for neutral cl

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop