The RDS-220 <span>hydrogen bomb, soviet </span>
Answer:
it just pulls them at the same time
Explanation:
Answer:
The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
Explanation:
Given;
initial temperature of the liquid, t₁ = 76.3 +/- 0.4⁰C
final temperature of the liquid, t₂ = 67.7 +/- 0.3⁰C
The change in temperature of the liquid is calculated as;
Δt = t₂ - t₁
Δt = (67.7 - 76.3) +/- (0.3 - 0.4)
Δt = (-8.6) +/- (-0.1)
Δt = 8.6 +/- 0.1 ⁰C
Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
Answer:
140 watt
Explanation:
We are given that
Force applied by student ,F=28 N
Weight pulled by students=70 N
Displacement,s=15 m
Time=3 s
We have to find the power developed by the student.
Work done=w=
Work done by the student=
Power=
Using the formula
Power=
Hence, the power developed by the students=140 watt
Answer:
b) 1.29 V
Explanation:
Potential Difference: This is the work done when one coulomb of charge moves from one point to another in an electric field
The expression for the potential difference is
Q = CV .................... Equation 1
Where Q = amount of charge, C = Capacitance of the capacitor, V = potential difference.
Making V the subject of the equation,
V = Q/C................. Equation 2
Given: Q = 4.5 µC = 4.5×10⁻⁶ C, C = 3.5 µF = 3.5×10⁻⁶ F
Substituting into equation 2
V = 4.5×10⁻⁶ /3.5×10⁻⁶
V = 1.2857 V
V ≈ 1.29 V.
Hence the right option is b) 1.29 V
