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lbvjy [14]
3 years ago
11

A runner is running to the left with a speed of 8.0\,\dfrac{\text m}{\text s}8.0 s m ​ 8, point, 0, start fraction, start text,

m, end text, divided by, start text, s, end text, end fraction when she speeds up with constant acceleration for 2.0\,\text s2.0s2, point, 0, start text, s, end text to reach the finish line. The runner's final speed is 9.9\,\dfrac{\text m}{\text s}9.9 s m ​ 9, point, 9, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the left. What was the runner's acceleration as she sped up to the finish line?
Physics
2 answers:
Alla [95]3 years ago
4 0

Answer:

The runner's acceleration as she sped up to the finish line is 0.95m/s²

Explanation:

Acceleration is the change in velocity of a body with respect to time. It is expressed as;

Acceleration = change in velocity/time

Change in velocity = final velocity - initial velocity

Acceleration = final velocity - initial velocity / time

Given initial speed = 8.0m/s

Final speed = 9.9m/s

Time taken = 2.0s

Acceleration = 9.9-8.0 / 2.0

Acceleration = 1.9/2

Acceleration = 0.95m/s²

yawa3891 [41]3 years ago
4 0

Answer:

-0.95

Explanation:

saw it on khan academy

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Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit
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Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

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Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, g=-9.81 m/s^2.

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v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4

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Here, the negative sign means the final velocity is in the downward direction.

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3 years ago
What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?
PtichkaEL [24]

Answer:

\tau=3.3*10^{-6}s

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

C*\frac{dV}{dt}+\frac{V}{R}=0

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

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\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:

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\tau=R*C=10*(0.330*10^{-6})=3.3*10^{-6}s

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Explanation:

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