Answer:
chloroplasts
Explanation:
Most plant shoots exhibit positive phototropism, and rearrange their chloroplasts in the leaves to maximize photosynthetic energy and promote growth.
Answer:
Technician B
Explanation:
here on analyzing both the statements from technician A and technician B. The Statement from Technician B is more logical and correct. That the power-assisted brake system reduces the force that the driver must exert on the brake pedal.
The power-assisted brake system does not reduce the distance of stopping. What it does is it reduces the force to be applied by the driver. Thus, making the drive more comfortable.
' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.
That's all the physics we need to know to answer this question.
The rest is just arithmetic.
(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)
= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)
= 51,840,000 joules
__________________________________
Wait a minute ! Hold up ! Hee haw ! Whoa !
Excuse me. That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's
(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)
= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)
= 14.4 kW·hour
Rounded to the nearest whole number:
14 kWh
I’m pretty sure it’s circuit three.
Answer:
6.67×10¯⁹ A
Explanation:
From the question given above, the following data were obtained:
Quantity of electricity (Q) = 2 μC
Time (t) = 5 mins
Current (I) =?
Next, we shall convert 2 μC to C. This can be obtained as follow:
1 μC = 1×10¯⁶ C
Therefore,
2 μC = 2 μC × 1×10¯⁶ C / 1 μC
2 μC = 2×10¯⁶ C
Next, we shall convert 5 mins to seconds. This can be obtained as follow:
1 min = 60 secs
Therefore,
5 min = 5 min × 60 sec / 1 min
5 mins = 300 s
Finally, we shall determine the current in the circuit. This can be obtained as follow:
Quantity of electricity (Q) = 2×10¯⁶ C
Time (t) = 300 s
Current (I) =?
Q = It
2×10¯⁶ = I × 300
Divide both side by 300
I = 2×10¯⁶ / 300
I = 6.67×10¯⁹ A
Thus, the current in the circuit is 6.67×10¯⁹ A
