Answer:
1.) 4m
2.) 37 m
3.) 62m
4.) 2.5 s
Explanation:
1.) Given that the
Thinking distance = 1m
Breaking distance = 3m
Stopping distance = breaking distance + thinking distance
Stopping distance = 1 + 3 = 4m
2.) Given that the
Stopping distance = 52 m
Thinking distance = 15m
Breaking distance = 52 - 15 = 37m
3.) The stopping distance = 76m
Thinking distance = 14m
Breaking distance = 76 - 14 = 62m
It take the brakes 62m to slow the car down to a stop.
4.) Given that a lorry travels 28m when stopping from a speed of 4m/s. If its braking distance was 18m, what was the driver’s reaction time?
Thinking = stopping distance - braking distance
Thinking distance = 28 - 18 = 10m
Speed = distance/time
4 = 10/reaction time
Reaction time = 10/4
Reaction time = 2.5 s
5.) Question incomplete
The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.
<h3>What is power of the circuit?</h3>
The power of the bulb or any resistor is equal to the product of voltage and current flowing through it.
P = VI
Circuit A has bulbs in series while the circuit B has bulbs in parallel.
When bulb 3 added to circuit A, the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.
The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.
Thus, the last option is correct.
Learn more about power.
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<span>The correct answer is (63.6, -13.2)
</span>
Ionization energy<span> is the </span>energy<span> required to remove an electron from a gaseous </span>atom<span> or ion.</span>
Answer:
Explanation:
give data:
inside diameter = 5.0 cm
charge q = 0.25 nC
Outside diameter = 15 cm
potential V at inside sphere is = 
potential V at outside sphere is = 
k is constant whose value is = 
then potential difference between two point is
![\Delta V = kq \left [\frac{1}{R}-\frac{1}{r} \right ]](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20kq%20%5Cleft%20%5B%5Cfrac%7B1%7D%7BR%7D-%5Cfrac%7B1%7D%7Br%7D%20%20%5Cright%20%5D)
![\Delta V = 9*10^{9}*0.25*10^{-9} \left [\frac{1}{0.05}-\frac{1}{0.15} \right ]](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%209%2A10%5E%7B9%7D%2A0.25%2A10%5E%7B-9%7D%20%5Cleft%20%5B%5Cfrac%7B1%7D%7B0.05%7D-%5Cfrac%7B1%7D%7B0.15%7D%20%5Cright%20%5D)
