17
What would the scale read? zero
18 In free fall you are being pulled by a gravity. "Truly" weightless presumably happens in deep space where there is nothing to pull you.
19 coasters accelerate down to simulate weight loss/zeroised. As do NASA planes,
Roller coasters are for fun seekers. NASA is for science
Answer:
N= 3
Explanation:
For this exercise we must use Faraday's law
E = - dФ / dt
Ф = B . A = B Acos θ
tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives
E = - A cos θ (B - B₀) / t
The angle enters the magnetic field and the normal to the area is zero
cos 0 = 1
A = π r²
In the length of the wire there are N turns each with a length L₀ = 2π r
L = N (2π r)
r = L / 2π N
we substitute
A = L² / (4π N²)
The magnetic field produced by a solenoid is
B = μ₀ N/L I
for which
B₀ = μ₀ N/L I
The final field is zero, because the current is zero
B = 0
We substitute
E = - (L² / 4π N²) (0 - μ₀ N/L I) / t
E = μ₀ L I / (4π N t)
N = μ₀ L I / (4π t E)
The electromotive force is E = 0.80 mV = 0.8 10⁻³ V
let's calculate
N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]
N = 320 10⁻⁷ / 9.6 10⁻⁶
N = 33.3 10⁻¹
N= 3
Answer:
Option A
You need a Angle C congruent to angle F
Explanation:
EX) Side angle Side = sas
Answer:
b) Nothing will happen, the sea saw will still be balanced.
Explanation:
b) Nothing will happen, the sea saw will still be balanced.
Reason:-
When two kids are balanced, the sum of torques on the seesaw will be zero.
if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.
Therefore the seesaw is balanced
Answer:
The energy stored in the solenoid is 7.078 x 10⁻⁵ J
Explanation:
Given;
diameter of the solenoid, d = 2.80 cm
radius of the solenoid, r = d/2 = 1.4 cm
length of the solenoid, L = 14 cm = 0.14 m
number of turns, N = 200 turns
current in the solenoid, I = 0.8 A
The cross sectional area of the solenoid is given as;
The inductance of the solenoid is given by;
The energy stored in the solenoid is given by;
E = ¹/₂LI²
E = ¹/₂(2.212 x 10⁻⁴)(0.8)²
E = 7.078 x 10⁻⁵ J
Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J
