Answer:
The rider's speed will be approximately 35 m/s
Explanation:
Initially the rider has kinetic and potential energy, and after going down the hill, some of the potencial energy turns into kinetic energy. So using the conservation of energy, we have that:
The kinetic and potencial energy are given by:
So we have that:
So the rider's speed will be approximately 35 m/s
Answer:
The final pressure of oxygen gas is 8.33 atm.
Explanation:
From the given data
V=26 m^3 or 26000 L
T1=270K
T2=440K
n1=6000 moles
R=0.0821 L.atm/molK
Now from the ideal gas equation
As the options given are not defined in which unit thus the final pressure of oxygen gas is 8.33 atm.
<em>*The options are provided for a different question where </em>
Answer:
5.3 x 10⁻⁹ C
Explanation:
r = radius of cylindrical shell = 10⁻⁵ m
L = length = 0.32 m
A = area
Area is given as
A = 2πrL
A = 2 (3.14) (10⁻⁵) (0.32)
A = 20.096 x 10⁻⁶ m²
d = separation = 10⁻⁸ m
= dielectric constant = 4
Capacitance is given as
eq-1
V = Potential difference across the membrane = 74 mV = 0.074 Volts
Q = magnitude of charge on each side
Magnitude of charge on each side is given as
Q = CV
using eq-1
Inserting the values
Q = 5.3 x 10⁻⁹ C