Answer:
The torque in the coil is 4.9 × 10⁻⁵ N.m
Explanation:
T = NIABsinθ
Where;
T is the torque on the coil
N is the number of loops = 9
I is the current = 7.8 A
A is the area of the circular coil = ?
B is the Earth's magnetic field = 5.5 × 10⁻⁵ T
θ is the angle of inclination = 90 - 56 = 34°
Area of the circular coil is calculated as follows;
![A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7D%20%5C%5C%5C%5CA%20%3D%20%5Cfrac%7B%5Cpi%200.17%5E2%7D%7B4%7D%20%3D0.0227%20m%5E2)
T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°
T = 4.9 × 10⁻⁵ N.m
Therefore, the torque in the coil is 4.9 × 10⁻⁵ N.m
The description of the problem is a bit ambiguous as "<span>increasing at the rate of </span>3 inches more than the time" can be interpreted in many ways. Assuming the increase rate is the (3+t) per second and as it expands for 3 seconds, then the diameter of the sphere would be:
d= (3 + t)*t
d= 3t + t^2= 18
r= 1/2 d
r= 1/2*18= 9
volume= 4/3 * pi * r^3
volume= 4/3 * pi * 9^3
volume= 4/3 * pi * 729= 972 pi cubic inches
Answer:
this effect is called SCREENING
Explanation:
When a positive charge is surrounded by several negative charges at a long distance, an apparent decrease in the positive charge is observed, this effect is called SCREENING and is noticeable in multielectronic atoms
This effect is responsible for the decrease in ionization energy in the larger atoms.
Answer:
No one is right
Explanation:
John Case:
The function
is defined between -1 and 1, So it is not possible obtain a value
greater.
In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.
Larry case:
Is you have
, the domain of this is [0,2].
it is equivalent to adding 1 to the domain of the
, and its mean that the function
, in general, is not greater than
.
Answer:
Explanation:
The gas law relation is
PV = nRT where n is no of moles of gas
given 1 kg of oxygen
n = 1000 / 32 = 31.25 moles
P = 30 bar = 30 x 10⁵ Pa
V =?
T = 180 K
R = 8.32
Substituting the values
30 x 10⁵ x V = 31.25 x 8.32 x 180
V = 1560 x 10⁻⁵ m³
= 1.56 x 10⁻² m³
When volume remains constant , pressure is proportional to temperature
P₂ / P₁ = T₂ / T₁
P₂ = ( 150 / 180 ) x 30
= 25 Bar.