Answer:
Final mass=0.89kg
Final pressure=5.6bar
Explanation:
To find mass,m=v/v1
But v1=vf + x(vg-vf)
Vf= 0.001093m^3/kg
Vg= 0.3748m^3/kg
V1= 0.001093+0.5(0.3748-0.001093)
V1= 0.225m^3/kg
M= 0.20/0.225 =0.89kg
Final pressure will be:
V/V1= P/P1
Cross multiply
VP1=V1P
P1= 0.225×5/0.2
P1=:5.6 bar
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a)
position of wire = 50 - 1.2
= 48.8cm
b)
Direction ⇒ downward
Answer:
Approximately 6.8 x 10⁻¹⁵
Explanation:
To be able to get this fraction, there are some things we need to know.
1. The radius of nucleus = 1.0 x 10⁻¹³ cm
2. The radius of hydrogen atom = 52.9 pm
3. Volume of sphere = V1/V2 = (R1/R2)^3
4. 1 picometer (pm) = 10^-10 cm
CHECK ATTACHMENT FOR Step by step solution to the answer
1. 1.59 s
The period of a pendulum is given by:
where L is the length of the pendulum and g the gravitational acceleration.
In this problem,
L = 0.625 m
g = 9.81 m/s^2
Substituting into the equation, we find
2. 54,340 oscillations
The total number of seconds in a day is given by:
So in order to find the number of oscillations of the pendulum in one day, we just need to divide the total number of seconds per day by the period of one oscillation:
3. 0.842 m
We want to increase the period of the pendulum by 16%, so the new period must be
Now we can re-arrange the equation for the period of the pendulum, using T=1.84 s, to find the new length of the pendulum that is required to produce this value of the period:
The answer is an ancient solar system