Answer:
This is because it steps up or steps down electrical voltage. It multiplies either voltage (if it is a voltage transformer )or current (if it is a current transformer), but it does not multiply electrical power.
Explanation:
A transformer steps up or steps down electrical voltage, by transmitting power at a voltage, V₁ and Current I₁ at one terminal, to a voltage, V₂ and Current I₂ at its other terminals, just like a lever transmits force from one point to another. Since the power transmitted remains the same, (energy per unit time remains constant), I₁V₁ = I₂V₂ ⇒ I₁/I₂ = V₂/V₁ = n (the turns ratio of the transformer). So, the turns ratio will determine if its a step-up or step-down transformer. V₂ = nV₁. So, if V₁ > V₂ it is a step down transformer and if V₁ < V₂ it is a step-up transformer.It multiplies either voltage (if it is a voltage transformer )or current (if it is a current transformer), but it does not multiply electrical power, since P = IV = constant for the transformer.
Is the sciencetjfic law and security theory the same?
TRUE
The product of an object's mass and velocity is B.momentum.
<span>♡♡Hope I helped!!! :)♡♡
</span>
Answer:
The answer to the question is
The ladybug begins to slide
Explanation:
To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same
Where the frictional force equals 
= μ×N = m×g×μ
and the centripetal force is given by m·ω²·r
If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have
m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁
and for the gentleman bug we have
m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂
But r₁ = 2×r₂
Therefore substituting the values of r₁ =2×r₂ we have
g×μ = ω²·r₁ = g×μ = ω²·2·r₂
Therefore ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide
The ladybug begins to slide
Answer:
metre for length and the kilogram for Mass
