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
Add 35 to 215. then divide by 25. you should get x=10
I think the answer would be letter B.
Answer:
A.) V = 14 m/s
B.) h = 36.6 m
Explanation:
Given the formula v = √2gh
where g = 9.8m/sec^2 is the acceleration due to gravity.
A.) Determine the impact velocity for an object dropped from a height of 10 m.
Substitute height h in the given formula
V = √2gh
V = √2 × 9.8 × 10
V = √196
V = 14 m/s
b. Determine the height required for an object to have an impact velocity of 26.8 m/sec (~ 60 mph). Round to the nearest tenth of a meter.
Substitute the velocity in the given formula and make height h the subject of formula.
26.8 = √2 × 9.8 × h
Square both sides
718.24 = 19.6h
h = 718.24 / 19.6
h = 36.64 m
h = 36.6 m
To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations
Work done to upward the object
Horizontal Force applied while carrying 10m,
Height descended in setting the child down
For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.
