The potential difference between points a and b is zero.
<h3>Total emf of the series circuit</h3>
The total emf in the circuit is the sum of all the emf in the circuit.
emf(total) = 1.5 + 1.5 = 3.0 V
<h3>Potential difference</h3>
The potential difference between two points, a and b is calculated as follows;
V(ab) = Va - Vb
V(ab) = 1.5 - 1.5
V(ab) = 0
Thus, the potential difference between points a and b is zero.
Learn more about potential difference here: brainly.com/question/3406867
Loss of habitats for fish, birds, and other wildlife. Sediment pollution is one of the leading causes of the loss of the wetlands, but it’s not just the wetlands. Changes in the nutrients in your water. The same problem that affects the fish in your area may also affect you. Other drinking water contamination.
Answer:
F = 520 N
Explanation:
For this exercise the rotational equilibrium equation should be used
Σ τ = 0
Let's set a reference system with the origin at the back of the refrigerator and the counterclockwise rotation as positive. On the x-axis it is horizontal directed outward, eg the horizontal y-axis directed to the side and the z-axis vertical
Torque is
τ = F x r
the bold indicate vectors, we analyze each force
the applied force is horizontal along the -x axis, the arm (perpendicular distance) is directed in the z axis,
The weight of the body is the vertical direction of the z-axis, so the arm is on the x-axis
-F z + W x = 0
F z = W x
F =
W
The exercise indicates the point of application of the force z = 1.5 m and the weight is placed in the center of mass of the body x = 0.6 m, we are assuming that the force is applied in the wide center of the refrigerator
let's calculate
F = 1300 0.6 / 1.5
F = 520 N
Its d all the above your welcome
Answer:
Explanation:
Let hotter star has surface area of A . The cooler star would have surface area 9 times that of hotter star ie 9A , because its radius is 3 times hot star. Let temperature of hot star be T ₁.
Total radiant energy is same for both the star
Using Stefan's formula of black body radiation,
For cold star E = 9A x σ T⁴
For hot star E = A x σ T₁⁴
A x σ T₁⁴ = 9A x σ T⁴
T₁⁴ = (√3)⁴T⁴
T₁ = √3T .
b )
Let the peak intensity wavelength be λ₁ and λ₂ for cold and hot star .
As per wein's law
for cold star , λ₁ T = b ( constant )
for hot star λ₂ √3T = b
dividing
λ₁ T / λ₂ √3T = 1
λ₂ / λ₁ = 1 / √3