882 divided by 9.81 (this is acceleration due to gravity) it equals 89.91
Answer:
can you put on a clearer image this one is hard to see
Answer:
51 Ω.
Explanation:
We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:
Resistor 1 (R₁) = 40 Ω
Resistor 3 (R₃) = 70.8 Ω
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?
Since the two resistors are in parallel connection, their equivalent can be obtained as follow:
R₁ₙ₃ = R₁ × R₃ / R₁ + R₃
R₁ₙ₃ = 40 × 70.8 / 40 + 70.8
R₁ₙ₃ = 2832 / 110.8
R₁ₙ₃ = 25.6 Ω
Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω
Resistor 2 (R₂) = 25.4 Ω
Equivalent Resistance (Rₑq) =?
Rₑq = R₁ₙ₃ + R₂ (series connection)
Rₑq = 25.6 + 25.4
Rₑq = 51 Ω
Therefore, the equivalent resistance of the group is 51 Ω.
Answer:
In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.
Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.
Explanation:
In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.
Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.