Answer:
C. Have your hazard lights on
Explanation:
Speeding up will cause an accident
Counter steering is not easy to do
Slowing down my result in you being rear ended
Handsaw teeth are very sharp: to avoid being cut by the teeth, keep hands and fingers well away from the
path of the blade
Answer:
The electric current from the batteries installed in a radio supplies direct current (DC) electricity to the radio components directly as an alternative source to the Alternating Current (AC) converted to DC by the power unit located at the radio end of the cable plugged into the wall outlet.
Explanation:
Part of the power unit in a radio includes an AC to DC converter, which is an electrical circuit that is able to convert the alternating current power input from the wall outlet into a direct current output to the radio with which the radio can work
The alternative source of electric current from the batteries installed in a radio bypasses the AC to DC converter and supplies power directly to the radio so it can also work.
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
Answer:
B. Acid rain.
C. Photochemical smog.
Explanation:
Oxides of nitrogen contribute to the formation of photochemical smog and acid rain. Photochemical smog is a type of smog produced when ultraviolet light from the sun reacts with nitrogen oxides in the atmosphere while on the other hand, when nitrogen oxide react with the water vapor in the atmosphere forming nitric acid which falls on the earth surface with the help of precipitation.