The distance between the two student having a mass of 73.5 Kg and 79.9 Kg with a force of attraction of 3.5×10¯⁸ N is 3.35 m
From the question given above, the following data were obtained:
Force of attraction (F) = 3.5×10¯⁸ N
Mass of 1st student (M₁) = 73.5 kg
Mass of 2n student (M₂) = 79.9 kg
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
<h3>Distance apart (r) =? </h3>
The distance apart of the two student can be obtained as illustrated below:
Cross multiply
3.5×10¯⁸ × r² = 3.9171×10¯⁷
Divide both side by 3.5×10¯⁸
Take the square root of both side
<h3>r = 3.35 m</h3>
Therefore, the distance between the two student is 3.35 m
Learn more: brainly.com/question/20856669
Answer:
the kinetic energy increases
Answer:
a) 2.1 × 10^-6 m b) 1.3 × 10^-4 m
Explanation:
The thermal expansion of super Invar 0.20 ×10^-6oC^-1 and the length of Super is 1.9 m
using the linear expansivity formula
ΔL = αLoΔt where ΔL is the change in length meters, α is the linear expansivity of Super Invar in oC^-1 and Δt is the change in temperature in oC. Substituting the value into the formula gives
ΔL = 0.2 × 10 ^-6 × 1.9 m ×5.5 = 2.1 × 10^-6 meters to two significant figure
b) the thermal expansion of steel is 1.3 × 10^-5 oC^-1 and the length of steel is 1.8 m
using the formula
ΔL of steel = 1.3 × 10^-5 × 1.8 × 5.5 = 1.3 × 10^-4 m to two significant figure.
Answer:
Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:
where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is
Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:
Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
We know v=u +at
Here u = 0
a= 1.5 units
t= 5 s
So v= 5×1.5 m/s = 7.5 m/s
