Arteries push blood to the rest of the body after leaving the heart.
Answer:
r=0.127
Explanation:
When connected in series
Current = I
When connected in parallel
Current = 10 I
We know that equivalent resistance
In series R = R₁+R₂
in parallel R= R₁R₂/(R₂+ R₁)
Given that voltage is constant (Vo)
V = I R
Vo = I (R₁+R₂) ------------1
Vo = 10 I (R₁R₂/(R₂+ R₁)) -------2
From above equations
10 I (R₁R₂/(R₂+ R₁)) = I (R₁+R₂)
10 R₁R₂ = (R₁+R₂) (R₂+ R₁)
10 R₁R₂ = 2 R₁R₂ + R₁² + R₂²
8 R₁R₂ = R₁² + R₂²
Given that
r = R₁/R₂
Divides by R₂²
8R₁/R₂ = ( R₁/R₂)²+ 1
8 r = r ² + 1
r ² - 8 r+ 1 =0
r= 0.127 and r= 7.87
But given that R₂>R₁ It means that r<1 only.
So the answer is r=0.127
Ek = 1/2 mv^2
9 × 10^4 = 1/2 × 800 × v^2
9 × 10^4/400 = 400 v^2 / 400
9 × 10^4/400 = v^2
√225 = v
15 ms⁻¹ = v
That's the only way I know how to work it out
I think in this case velocity and speed would be considered the same because me
s = d/t and v=d/t
one is distance travelled and the other is displacement of a body
Answer:
(a) 43.2 kC
(b) 0.012V kWh
(c) 0.108V cents
Explanation:
<u>Given:</u>
- i = current flow = 3 A
- t = time interval for which the current flow =
- V = terminal voltage of the battery
- R = rate of energy = 9 cents/kWh
<u>Assume:</u>
- Q = charge transported as a result of charging
- E = energy expended
- C = cost of charging
Part (a):
We know that the charge flow rate is the electric current flow through a wire.
Hence, 43.2 kC of charge is transported as a result of charging.
Part (b):
We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:
Hence, 0.012V kWh is expended in charging the battery.
Part (c):
We know that the energy cost is equal to the product of energy expended and the rate of energy.
Hence, 0.108V cents is the charging cost of the battery.
Answer: exotic medium can be engineered to slow light
Explanation:
