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Grace [21]
3 years ago
13

A particle moving in a straight line with a constant acceleration covers 10m in 2 seconds and 22m in a further 2 seconds. How mu

ch further does it travel in the next two seconds?
Physics
1 answer:
MrRissso [65]3 years ago
7 0

The particle will travel  a distance of 28 m in the next two seconds

<h3>Acceleration</h3>

Acceleration is defined as the change in velocity with time

  • Acceleration = final velocity - initial velocity / time
<h3>Calculating initial velocity</h3>

distance covered = 10 m

time = 2 seconds

velocity = distance/time

initial velocity = 10/2

initial velocity = 5m/s

<h3>Calculating final velocity</h3>

distance = 22 m

time = 2 seconds

Final velocity = 22/2

Final velocity = 11 m/s

<h3>Calculating acceleration</h3>

Acceleration = change in velocity/time

acceleration = 11m/s - 5m/s /2 s

acceleration = 3m/s²

<h3>Calculating distance covered</h3>

Using the equation of motion:

s = ut + at²/2

s = 11 * 2 + 3 * 2²/2

s -=28 m

Therefore, the particle will travel  a distance of 28 m in the next two seconds

Learn more about distance and acceleration at: brainly.com/question/16847527

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The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You
tatiyna

Answer:

Explanation:

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You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
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Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
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2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
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3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
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