Answer:
1) R1 + ((R2 × R3)/(R2 + R3))
2) 0.5 A
3) 3.6 V
Explanation:
1) We can see that resistors R2 and R3 are in parallel.
Formula for sum of parallel resistors; 1/Rt = 1/R2 + 1/R3
Making Rt the subject gives;
Rt = (R2 × R3)/(R2 + R3)
Now, Resistor R1 is in series with this sum of R2 and R3. Thus;
Total resistance of circuit = R1 + ((R2 × R3)/(R2 + R3))
2) R_total = R1 + ((R2 × R3)/(R2 + R3))
We are given;
R1 = 7.2 Ω
R2 = 8 Ω
R3 = 12 Ω
R_total = 7.2 + ((8 × 12)/(8 + 12))
R_total = 7.2 + 4.8
R_total = 12 Ω
Formula for current is;
I = V/R
I = 6/12
I = 0.5 A
3) since current through the circuit is 0.5 and R1 is 7.2 Ω.
Thus, potential difference through R1 is;
V = IR = 0.5 × 7.2 = 3.6 V
Answer:
T=3.29 N
Explanation:
let V be the speed at low point
From conservation of energy
0.5mV²=m*g*l*(1-cos(∅))
(mV²/l)=2m*g*(1-cos(∅))
Nowat low point
T-mg=mV²/l
T=mg+2mg((1-cos(∅))
T=mg(3-2cos(∅))
T=0.25*.81*(3-2cos(34))
T=3.29 N
Rise over run at 1 second
It’s the same slope from 0 to 2 seconds
10/2=5mps
As a note all time points between 0and 2 will have this instantaneous velocity
Instantaneous velocity at time 2 is 0
Answer:
7650 m.
Explanation:
Ocean floor depth, d is:
d = v * t,
where,
d = the distance from the vessel to the ocean floor (or the depth)
v = 1530 m/s = velocity of the ultrasonic sound
t = t_echo/2 = time that the ultrasonic sound needs to reach the ocean floor
t_echo = 10 s = time that the ultrasonic sound needs to reach the ocean floor and return back to the vessel.
d = v * t
= v * t_echo/2
= 1530 * 10/2
= 7650 m.