Horizontal component = (10N) · sin (20°) = 3.42... N (rounded)
Vertical component = (10N) · cos (20°) = 9.39... N (rounded)
Answer:
0.0000045 s
Explanation:
f = Frequency = 8 MHz
Clock cycle is given by

Time taken for 12 clock cycles

Time taken per instruction is 0.0000015 s
In reading and displaying information it requires 3 processes
1 for reading, 1 for searching and 1 for displaying.

Time taken is 0.0000045 s
Answer:
37357 sec
or 622 min
or 10.4 hrs
Explanation:
GIVEN DATA:
Lifting weight 80 kg
1 cal = 4184 J
from information given in question we have
one lb fat consist of 3500 calories = 3500 x 4184 J
= 14.644 x 10^6 J
Energy burns in 1 lift = m g h
= 80 x 9.8 x 1 = 784 J
lifts required 
= 18679
from the question,
1 lift in 2 sec.
so, total time = 18679 x 2 = 37357 sec
or 622 min
or 10.4 hrs
Amanda might be suffering from a disease called Sickle Cell Anemia. It is an inherited red blood cell disorder.
Answer:
Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.
.
Explanation:
Consider four possible cases.
<h3>Case A: 12.0 V.</h3>

In case all three capacitors are connected in parallel, the
capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.
<h3>Case B: 5.54 V.</h3>
![-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-](https://tex.z-dn.net/?f=-3.0%5C%3B%5Cmu%5Ctext%7BF%7D-%5B%5Cbegin%7Barray%7D%7Bc%7D-%7B%5Cbf%202.0%5C%3B%5Cmu%5Ctext%7BF%7D%7D-%5C%5C-1.5%5C%3B%5Cmu%5Ctext%7BF%7D-%5Cend%7Barray%7D%5D-)
In case the
capacitor is connected in parallel with the
capacitor, and the two capacitors in parallel is connected to the
capacitor in series.
The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.
The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,
.
What will be the voltage across the 2.0 μF capacitor?
The charge stored in two capacitors in series is the same as the charge in each capacitor.
.
Voltage is the same across two capacitors in parallel.As a result,
.
<h3>Case C: 2.76 V.</h3>
.
Similarly,
- the effective capacitance of the two capacitors in parallel is 5.0 μF;
- the effective capacitance of the three capacitors, combined:
.
Charge stored:
.
Voltage:
.
<h3 /><h3>Case D: 4.00 V</h3>
.
Connect all three capacitors in series.
.
For each of the three capacitors:
.
For the
capacitor:
.