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:
. Radio waves, infrared rays, visible light, ultraviolet rays, X-rays, and gamma rays are all types of electromagnetic radiation.
Explanation:
Radio waves have the longest wavelength, and gamma rays have the shortest wavelength.
Answer:
A) The particle will accelerate in the direction of point C.
Explanation:
As we know that
potential at points A, B,C and D as V_A, V_B, V_C, V_D and it is clear from the question that
V_A>V_B>V_C
And we know that flow is always from higher to lower potential (for positive charge due to positive potential energy).
So the charge will accelerate from B toward C.
Hence, the correct option is A.
The kinetic energy gained by the air molecules is 0.0437 J
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Given:
Mass of a coffee filter, m = 1.5 g
Height from which it is dropped, h = 3 m
Speed at ground, v = 0.7 m/s
Initially, the coffee filter has potential energy. It is given by :
P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m
P = 0.0441 J
Finally, it will have kinetic energy. It is given by :
×
× 10⁻³ × (0.7)²
E = 0.000343 J
The kinetic energy Kair did the air molecules gain from the falling coffee filter is :
E = 0.000343 - 0.0441
= 0.0437 J
So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J
Learn more about kinetic energy here:
brainly.com/question/8101588
#SPJ4
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
