a=v^2/r. This is the centripetal acceleration for object moving in circle. In order to imitate the earths gravitational acceleration, plug in 9.8 for a. Then plug in the radius of the cylinder and solve for v.
Answer:
Density = 3 x 10⁻⁵ kg/m³
Explanation:
First, we will find the volume of the planet:
Now, we will use the expression for gravitational force to find the mass of the planet:
where,
m = mass = ?
g = acceleration due to gravity = 6.67 x 10⁻¹¹ m/s²
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²
r = radius = 8000 m
Therefore,
Therefore, the density will be:
<u>Density = 3 x 10⁻⁵ kg/m³</u>
Answer:
330.4 N and 
counterclocwise to the x direction
Explanation:
Sum of forces in vertical are equal, let movement to right and upwards be positive while left and downwards be negative
Net force in horizontal direction is 85-115=-30 N
Net force in vertical direction is 565-236=329 N
Resultant force=
Direction=
180-84.8=95.2
Therefore, direction is counterclocwise to the x direction
In a solid particles vibrate back and forth
Answer:
a) Q1= Q2= 11.75×10^-6Coulombs
b) Q1 =15×10^-6coulombs
Q2 = 38.75×10^-6coulombs
Explanation:
a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as
1/Ct = 1/C1 + 1/C2
Given C1 = 3.00 μF C2 = 7.75μF
1/Ct = 1/3+1/7.73
1/Ct = 0.333+ 0.129
1/Ct = 0.462
Ct = 1/0.462
Ct = 2.35μF
V = 5.00Volts
To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance
Q = 2.35×10^-6× 5
Q = 11.75×10^-6Coulombs
Since same charge flows through a series connected capacitors, therefore Q1= Q2=
11.75×10^-6Coulombs
b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2
C = 3.00 μF + 7.75 μF
C = 10.75 μF
For 3.00 μF capacitance, the charge on it will be Q1 = C1V
Q1 = 3×10^-6 × 5
Q1 =15×10^-6coulombs
For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5
Q2 = 38.75×10^-6coulombs
Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.
