Answer:
Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Explanation:
Given;
number of turns of solenoid, N = 269 turn
length of the solenoid, L = 102 cm = 1.02 m
radius of the solenoid, r = 2.3 cm = 0.023 m
current in the solenoid, I = 3.9 A
Magnitude of the magnetic field inside the solenoid near its centre is calculated as;
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Answer:
Stress = F / A force per unit area
A = 3.00 cm^2 = 3 E-4 m^2
F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied
F/3 = 2.4E4 N if force not to exceed limit (= f)
f = M a
a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g
Answer:
F=X.F=mxq. 3. 1 N/kg=0.5kg g=9.80
When t=2, the ball has fallen d(2) = 16 (2²) = 64 feet .
When t=5, the ball has fallen d(5) = 16 (5²) = 400 feet .
Distance fallen from t=2 until t=5 is (400 - 64) = 336 feet.
Time period between t=2 until t=5 is (5 - 2) = 3 seconds.
Average speed of the ball from t=2 until t=5 is
(distance covered) / (time to cover the distance)
= 336 feet / 3 seconds = 112 feet per second.
That's what choice-C says.
Answer:
The function that describe the motion in the time
y (t) = 0.28m * sin ( 36.025 * t)
Explanation:
The angular frequency of oscillation of the spring
w = √k/m
w = √305 N/m / 0.235 kg
w = 36.025 rad / s
To determine the function of the motion knowing as a motion oscillation in a amplitude a frequency
y(t) = A * sin (w t )
So
A = 28.0 cm * 1 m / 100 cm = 0.28 m
So replacing to determine the function of the motion in the time
y (t) = A sin (w t)
y (t) = 0.28m * sin ( 36.025 * t)
