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meriva
3 years ago
9

An electric vehicle starts from rest and accelerates at a rate of 2.3 m/s2 in a straight line until it reaches a speed of 29 m/s

. The vehicle then slows at a constant rate of 1.5 m/s2 until it stops. (a) How much time elapses from start to stop
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0

Answer:

t = 12.6 seconds

Explanation:

Given that,

Initial velocity, u = 0

Acceleration of an electric vehicle, a = 2.3 m/s²

Final velocity, v = 29 m/s

We need to find the time elapses from start to stop. The acceleration of an object is given by the relation as follows :

a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{29-0}{2.3}\\\\t=12.6\ s

So, 12.6 seconds is elapsed from start to stop.

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Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm.
pantera1 [17]

Answer:

R_a/R_b=3

Explanation:

The resistance in terms of the area and the length of the wire is given by:

R=pL/A

if we have two wires, the first one is a solid wire with a diameter of d_A = 1 x 10^-3 m, and the second one is a hollow wire with inner diameter of d_B,i = 1 x 10^-3 m and outer diameter of d_B,σ= 2 x 10^-3 m, so the cross sectional area of the first wire is:  

A_a=πr^2_a

A_a=πd^2_A/4

hence the resistance is:  

R_a=(4*p*L_a)/π*d^2_A                                     (1)

the area of the second wire is:  

A_b=π*r^2_B,σ-π*r^2_B,i

A_b=π/4(d^2_B,σ-d^2_B,i)

hence the resistance is:  

R_b=(4*p*L_b)/π(d^2_B,σ-d^2_B,i)                   (2)

To find the ratio between the resistances R_a/R_b, we divide (1) over (2) to get:  

R_a/R_b=(d^2_B,σ-d^2_B,i)*L_a/(d^2_a*L_b)

but the wires have the same length, therefore:  

R_a/R_b=(d^2_B,σ-d^2_B,i)/(d^2_a)

substitute with the given values to get:

R_a/R_b=3

6 0
3 years ago
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Explanation:

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sergejj [24]
I don’t know how air would be an insulator so I’m guessing that one isn’t
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Answer: The correct answer is option (C).

Explanation:

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