The engine in an imaginary sportThe engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). At full power, the car can accelerate from zero to 30.0 mph in time 1.00 s .s car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). Required:
a. At full power, how long would it take for the car to accelerate from 0 to 58.0mph?
b. A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?

a. 1.93 s b. 3 s

Explanation:

a. At full power, how long would it take for the car to accelerate from 0 to 58.0mph?

Since the car accelerates from 0 to 30 mph in 1.00 s, we find its acceleration, a from a = (v - u)/t where u = 0 m/s, v = 30 mph and t = 1.00s = 1/3600 h

So, substituting the values of the variables into the equation, we have

a = (v - u)/t

a = (30 mph - 0 mph)/ 1/3600 h

a = 30 mph × 3600 /h

a = 108000 mph²

So, we find the time it takes the car to accelerate to 58 mph from 0 mph from

t = (v' - u')/a where u = 0 mph, v = 58 mph and a = 108000 mph²

So, substituting the value of the variables into the equation, we have

t = (v' - u')/a

t = (58 mph - 0 mph)/108000 mph²

t = 58 mph/108000 mph²

t = 5.37 × 10⁻⁴ h

t = 5.37 × 10⁻⁴ × 3600 s

t = 1.93 s

b. A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?

Since the car accelerates from 0 to 29 mph in 1.50 s, we find its acceleration, a from a = (v - u)/t where u = 0 m/s, v = 29 mph and t = 1.05s = 1/3600 h

So, substituting the values of the variables into the equation, we have

a = (v - u)/t

a = (29 mph - 0 mph)/ 1.5/3600 h

a = 29 mph × 3600/1.5 /h

a = 104400/1.5 mph²

a = 69600 mph²

So, we find the time it takes the car to accelerate to 58 mph from 0 mph from

t = (v' - u')/a where u = 0 mph, v = 58 mph and a = 69600 mph²

So, substituting the value of the variables into the equation, we have

t = (v' - u')/a

t = (58 mph - 0 mph)/69600 mph²

t = 58 mph/69600 mph²

t = 8.33 × 10⁻⁴ h

t = 8.33 × 10⁻⁴ × 3600 s

t = 3 s

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

well clustered together.

points close on a scatter diagram