When you drop a 0.43 kg apple, earth exerts
a force on it that accelerates it at 9.8 m/s^2
toward the earth’s surface. according to newton’s third law, the apple must exert an equal
but opposite force on earth.
if the mass of the earth 5.98 × 1024 kg, what
is the magnitude of the earth’s acceleration
toward the apple?
answer in units of m/s^2

The acceleration of the earth is 7.05 * 10^-25 m/s²

Explanation:

Step 1: Data given

mass of the apple = 0.43 kg

acceleration = 9.8 m/s²

mass of earth = 5.98 * 10 ^24 kg

Step 2: Calculate the acceleration of the earth

Following the third law of Newton F = m*a

F(apple) = F(earth) = m(apple)*a(apple)

F(apple) = 0.43 kg * 9.8 m/s² = 4.214 N

a(earth) = F(apple/earth)/m(earth)

a(earth) = 4.214N /5.98 * 10 ^24 kg

a(earth) = 7.05 * 10^-25 m/s²

The acceleration of the earth is 7.05 * 10^-25 m/s²

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: