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3 years ago

12

A driver is moving at 17.0 m/s when she sees a child on a bike 65m ahead of her. If the car can brake at 4.9 m/s^2 and the drive

rs reaction time is 1.6 s, can she avoid hitting the child

1 answer:

Lera25 [3.4K]3 years ago

7 0

Answer:

The driver can avoid the child because at the time she brake 1.6s the car just move 20.928 meters so is far to the 65 meters where the kind is

Distance= 20.928 m

Explanation:

$V_{o} = 17 \frac{m}{s} \\a= -4.9 \frac{m}{s^{2} }\\ t= 1.6 s$

Distance the kid is 65m so the car have to be less in the time she is braking

So to calculated the distance while she is braking

$X_{f} = X_{o} +V_{o}*t + \frac{1}{2} * a *t^{2}\\ X_{f} = 0m +17 \frac{m}{s} *1.6 s + \frac{1}{2} * 4.9 \frac{m}{s^{2} } *(1.6s)^{2}\\X_{f} = 27.2 m+ (-6.272 m)\\X_{f} = 20.928 m$

The distance of the car is less than the distance of the kid in his bike

So she didn't hit him

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At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

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(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

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Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

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(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

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$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

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$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

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After several seconds,

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If we use this approximation into the formula

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We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

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