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torisob [31]
4 years ago
6

A sport car moving at constant speed travels 110m in 5.0 s. if it then brakes and comes to a stop in 4.0 s, what is the magnitud

e of its acceleration in m/s2, and in g’s (g=9.80 m/s2)? 2
Physics
1 answer:
stiks02 [169]4 years ago
7 0
D = 110 m,  t = 5 s
v o = 110 cs : 5 m = 22 m/s
-------------------------------------
v = v o - a t
v = 0 m/s,  v o =  22 m/s,  t = 4 s
0 = 22 - 4 a
4 a = 22
a = 22 : 4
a = 5.5 m/s²
g = 9.80 m/s²
9.80 : 5.5 = 0.56 
Answer:
The magnitude of its acceleration is 5.5 m/s or 0.56 g. 
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Answer:

Part A 2.88 A, PART B 28.82 mA PART C  0.288 mA

Explanation:

We have given angular frequency \omega =850rad/sec

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Part A

We have given inductance L=10^{-2}H

So inductive reactance X_L=\omega L=850\times 10^{-2}=8.5ohm

So current i=\frac{V}{X_L}=\frac{24.5}{8.5}=2.8823A

Part B

We have given inductance L=1 H

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Part C

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3 years ago
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PtichkaEL [24]

Answer:

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This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

\frac{dV}{V}=-\frac{1}{RC}dt

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt

Evaluating the integrals:

ln(\frac{V}{v})=e^{\frac{-t}{RC} }

natural logarithm to both sides in order to isolate V:

V(t)=ve^{-\frac{t}{RC} }

Where the term RC is called time constant and is given by:

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