KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC, is on the x axis at x=0.500 m.
No, there is no point at a finite distance where the electric potential is zero.
Hence, Option D) is correct.
What is electric potential?
Electric potential is the capacity for doing work. In the electrical case, a charge will exert a force on some other charge and the potential energy arises. For example, if a positive charge Q is fixed at some point in space, any other positive charge when brought close to it will experience a repulsive force and will therefore have potential energy.
It is also defined as the amount of work required to move a unit charge from a reference point to a specific point against an electric field.
To learn more about electric potential, refer to:
brainly.com/question/15764612
#SPJ4
Answer:
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Explanation:given values
Half life of lipase t_1/2 = 8 min x 60s/min = 480 s
Rate constant for first order reaction
k_d = 0.6932/480 = 1.44 x 10^-3 s-1
Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L
rate of hydrolysis Vm0 = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3
K_m = 5mmol/L
time take is given by
![t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]](https://tex.z-dn.net/?f=t%3D%20-%5Cfrac%7B1%7D%7BK_d%7Dln%5B1-%5Cfrac%7BK_d%7D%7BV_m_0%7D%28k_mln%5Cfrac%7Bs_0%7D%7Bs%7D%2B%28s_0-s%29%29%5D)
all values are given and putting these value we get
t=1642.83 secs
which is equal to
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Statements 1, 3, and 5 are true.
(A, C, and E)
