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tangare [24]
1 year ago
14

You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that

your reaction time is 0.50s and that your car brakes with constant acceleration. How far are you from the intersection when you begin to apply the brakes? What acceleration will bring you to rest right at the intersection? How long does it take you to stop after the light changes? the known values: vi=20m/s xi=110m t=.50s vf=0m/s xf=0m a=?
Physics
1 answer:
oksano4ka [1.4K]1 year ago
3 0

As we know that reaction time will be

t = 0.50 s

so the distance moved by car in reaction time

d = vt

d = 20 \times 0.50

d = 10 m

now the distance remain after that from intersection point is given by

d = 110 - 10 = 100 m

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

v_f = 0

v_i = 20 m/s

now from kinematics equation we will have

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2(a)100

a = \frac{-400}{200} = -2 m/s^2

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

v_f - v_i = at

0 - 20 = -2 (t)

t = 10 s

total time to stop the car is given as

T = 10 s + 0.5 s = 10.5 s

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The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the
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q = 4.5 nC

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In a series circuit, a generator (1300 Hz, 12.0 V) is connected to a 14.0- resistor, a 4.40-μF capacitor, and a 6.00-mH inductor
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Answer:

(a) 2.8 V

(b) 5.6 V

(c) 9.8 V

Explanation:

Given:

Frequency of the generator (f) = 1300 Hz)

Terminal voltage (V) =12.0 V

Resistance of resistor (R) = 14.0 Ω

Capacitance of capacitor (C) = 4.40 μF = 4.40 × 10⁻⁶ F

Inductance of the inductor (L) = 6.00 mH = 6.00 × 10⁻³ H

In order to find the voltages across each, we first need to find the reactance and impedance.

Reactance of the inductor is given as:

X_L=2\pi f L\\\\X_L=2\times 3.14\times 1300\times 6.00\times 10^{-3}\\\\X_L=49\ \Omega

Reactance of the capacitor is given as:

X_C=\frac{1}{2\pi fC}\\\\X_C=\frac{1}{2\times 3.14\times 1300\times 4.40\times 10^{-6}}\\\\X_C =28\ \Omega

Now, impedance is given as:

Z=\sqrt{X_L^2+X_C^2}\\\\Z=\sqrt{(49)^2+(28)^2}\\\\Z=\sqrt{3185}=56.4\ \Omega

Current across the circuit is given as:

I=\frac{V}{Z}\\\\I=\frac{12}{56.4}=0.2\ A

As resistor, capacitor and inductor are connected in series, the current across each of them is same and equal to total current in the circuit.

(a)

Voltage across the resistor is given as:

V_R=IR\\\\V_R=0.2\times 14=2.8\ V

Therefore, the voltage across resistor is 2.8 V.

(b)

Voltage across the capacitor is given as:

V_C=IX_C\\\\V_C=0.2\times 28=5.6\ V

Therefore, the voltage across the capacitor is 5.6 V.

(c)

Voltage across the inductor is given as:

V_L=IX_L\\\\V_L=0.2\times 49=9.8\ V

Therefore, the voltage across the inductor is 9.8 V.

6 0
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