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Gre4nikov [31]
3 years ago
8

When a 1.0-kilogram cart moving with a speed of

Physics
2 answers:
sp2606 [1]3 years ago
5 0
Hello there.

<span>When a 1.0-kilogram cart moving with a speed of
0.50 meter per second on a horizontal surface
collides with a second 1.0-kilogram cart initially
at rest, the carts lock together. What is the speed
of the combined carts after the collision?

</span><span>(3) 0.25 m/s</span>
neonofarm [45]3 years ago
4 0
The type of collision in this system is called inelastic collision where after collision the two objects stick together. We can solve for the final velocity after collision by the expression:

v = m1v1 + m2V2 / (m1+m2)

v = 1(0.50) + (1)(0) / (1+1) = 0.25 m/s 

Therefore, option 3 is the correct answer.
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Answer:

a) R₂/R₁ = 0.3049

b) I₂ / I₁  =  1.81

c) (230)² / ( 127)² =  3.28

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By definition  the power consumption is:

P = I²*R    ⇒  as  I = V/R   I² = (V/R)²    and  P = (V²/R²)*R

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I = current

R = is the resistor

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P₁  = V₁² /R₁     P₁  =  (127)²/ R₁

Case Device 2 Voltage  230 (V)

P₂  = ( 230)²/ R₂

As   P₁  =  P₂        (127)²/R₁  =  (230)²/R₂

(127)²/(230)²  =  R₂/R₁

a) R₂/R₁ = 0.3049

b)  P₁ = I₁²*R₁              P₂  =  I₂²*R₂

I₂² *R₂  =  I₁²*R₁

I₂² / I₁²   =  R₁/R₂

I₂² / I₁²   =  1/ 0.3049

I₂² / I₁²   =  3.2797

√  (I₂² / I₁² )   =  √3.2797

b) I₂ / I₁  =  1.81

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P₁´  =  V₂² / R₁

P₁  =  (127)²/R₁      and     P₁´ = (230)²/ R₁

Then the increasing factor is:

c) (230)² / ( 127)² =  3.28

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