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10 months ago

11

a car initially at rest can accelerate at 7 m/s^2 how long will it take the car to reach 60 m/s and how far will it travel durin

g this time

1 answer:

Alex787 [66]10 months ago

7 0

1. The time taken for the car to reach a velocity of 60 m/s is 8.57 s

2. The distance travelled during the time is 257.14 m

<h3>What is acceleration? </h3>

The acceleration of an object is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration
v is the final velocity
u is the initial velocity
t is the time

1. How to determine the time

Initial velocity (u) = 0 m/s
Acceleration (a) = 7 m/s²
Final velocity (v) = 60 m/s
Time (t) =?

a = (v – u) / t

Thus,

t = (v – u) / a

t = (60 – 0) / 7

t = 8.57 s

2. How to determine the distance

Initial velocity (u) = 0 m/s
Acceleration (a) = 7 m/s²
Final velocity (v) = 60 m/s
Distance (s) = ?

v² = u² + 2as

60² = 0² + (2 × 7 × s)

3600 = 0 + 14s

3600 = 14s

Divide both sides by 14

s = 3600 / 14

s = 257.14 m

Learn more about acceleration and velocity:

brainly.com/question/491732

brainly.com/question/19466392

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The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

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We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

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$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

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$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

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Answer:

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