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:
Hey there!
Stopwatch X recorded 40 seconds, and stopwatch Y recorded 50 seconds.
Stopwatch Y recorded 10 seconds longer than stopwatch X.
Hope this helps :)
Answer:
the potential energy is 114 J.
Explanation:
Given;
total mechanical energy, E = 400 J
kinetic energy, K.E = 286 J
The potential energy is calculated as follows;
E = K.E + P.E
where;
P.E is the potential energy
P.E = E - K.E
P.E = 400 J - 286 J
P.E = 114 J
Therefore, the potential energy is 114 J.
Answer:
h = 23.716 m
Explanation:
Given that,
The time taken by the stone to hit the water is, t = 2.2 s
Height of the bridge above the ground, h = ?
The distance that the body will fall through the time is given by the formula
S = 1/2 gt² m
Where,
g - acceleration due to gravity
Substituting the values in the above equation
S = 1/2 x 9.8 m/s² x (2.2 s)²
= 23.716 m
Therefore, the height of the bridge from the surface of the water is h = 23.716 m
(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.
(b) The acceleration of the child-wagon system is 0.33 m/s².
(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.
<h3>
Net force on the third child</h3>
Apply Newton's second law of motion;
∑F = ma
where;
- ∑F is net force
- m is mass of the third child
- a is acceleration of the third child
∑F = 96 N - 75 N - 12 N = 9 N
Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;
- the wagon
- the children outside the wagon
<h3>Free body diagram</h3>
→ → Ф ←
1st child friction wagon 2nd child
<h3>Acceleration of the child and wagon system</h3>
a = ∑F/m
a = 9 N / 27 kg
a = 0.33 m/s²
<h3>When the frictional force is 21 N</h3>
∑F = 96 N - 75 N - 21 N = 0 N
a = ∑F/m
a = 0/27 kg
a = 0 m/s²
Learn more about net force here: brainly.com/question/14361879
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