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LuckyWell [14K]
4 years ago
9

Plzz help me with science

Physics
1 answer:
RoseWind [281]4 years ago
3 0
These are the answers, i am not sure for the first question ( sry) :
2) D
3) D
5)A
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The blood vessels that carry blood away from the heart to the rest of the body are
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Arteries push blood to the rest of the body after leaving the heart.
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3 years ago
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Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are
Contact [7]

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

8 0
4 years ago
What is the speed of an 800 kg automobile if it has a kinetic energy of 9.00 x 10^J?
MA_775_DIABLO [31]

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

7 0
2 years ago
Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
kirza4 [7]

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 = 4\ h = 4\times 3600\ s = 14400\ s
  • 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.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

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:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

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.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

4 0
4 years ago
What machine would benefit from slowing lights speed
VARVARA [1.3K]

Answer: exotic medium can be engineered to slow light

Explanation:

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