Answer:
Output voltage is 1.507 mV
Solution:
As per the question:
Nominal resistance, R = 
Fixed resistance, R =
Gauge Factor, G.F = 2.01
Supply Voltage, 
Strain, 
Now,
To calculate the output voltage, 
:
WE know that strain is given by:
Thus
Now, substituting the suitable values in the above eqn:
<span>The answer is d. they are defined by their different physical features. A soil horizon is a layer in a soil profile. Horizons are defined by different physical features such as colour or texture. Other features include, structure and permeability. Each horizon is represented by a letter; A,B,C....</span>
Answer:
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Explanation:
The distance travelled on the rough ice is equal to the width of the rough ice.
distance d = 5.0 m
Initial speed u = 9.2 m/s
Final speed v = 5.8 m/s
The time taken to move through the rough ice can be calculated using the equation of motion;
d = 0.5(u+v)t
time t = 2d/(u+v)
Substituting the given values;
t = 2(5)/(9.2+5.8)
t = 2/3 = 0.66667 second
The acceleration is the change in velocity per unit time;
acceleration a = ∆v/t
a = (v-u)/t
Substituting the values;
a = (5.8-9.2)/0.66667
a = -5.099974500127
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Answer:
Fnet = F√2
Fnet = kq²/r² √2
Explanation:
A exerts a force F on B, and C exerts an equal force F on B perpendicular to that. The net force can be found with Pythagorean theorem:
Fnet = √(F² + F²)
Fnet = F√2
The force between two charges particles is:
F = k q₁ q₂ / r²
where
k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.
If we say the charge of each particle is q, then:
F = kq²/r²
Substituting:
Fnet = kq²/r² √2
Answer: sheet of charge
Explanation:
a )
Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.
b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.
C ) Electric field which is almost uniform near the sheet of charge is equal t the following
E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²
E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²
= .3389 x 10³
= 338.9 V / m
spacing between 1 V
= 1 / 338.9 m
= 2.95 X 10⁻3 m
= 2.95 mm.
