The condition is a neuron in when the outside of the neuron has a net positive charge and the inside has a net negative charge (due to accumulation of more sodium ions) is C. resting potential. T<span>he </span>resting membrane<span> </span>potential<span> of a </span>neuron<span> is approximately -70 mV (mV=</span><span>millivolt)</span>
Answer:
Vf= 7.29 m/s
Explanation:
Two force act on the object:
1) Gravity
2) Air resistance
Upward motion:
Initial velocity = Vi= 10 m/s
Final velocity = Vf= 0 m/s
Gravity acting downward = g = -9.8 m/s²
Air resistance acting downward = a₁ = - 3 m/s²
Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²
( Acceleration is consider negative if it is in opposite direction of velocity )
Now
2as = Vf² - Vi²
⇒ 2 * (-12.8) *s = 0 - 10²
⇒-25.6 *s = -100
⇒ s = 100/ 25.6
⇒ s = 3.9 m
Downward motion:
Vi= 0 m/s
s = 3.9 m
Gravity acting downward = g = 9.8 m/s²
Air resistance acting upward = a₁ = - 3 m/s²
Net acceleration = a = g - a₁ = 9.8 - 3 = 6.8 m/s²
Now
2as = Vf² - Vi²
⇒ 2 * 6.8 * 3.9 = Vf² - 0
⇒ Vf² = 53. 125
⇒ Vf= 7.29 m/s
Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,

Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?

a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)

s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Answer:
a= 4.4×10 m/s^2
Explanation:
pressure P = E/c
Where, E = 100 W/m^2 intensity of light
c= speed of light = 3×10^8 m/s
P = 1000/ 3×10^8
P = 3.33×10^(-6) Pa
Force F = P×A
- P is the pressure and c= speed of light
F = 3.33×10^{-6}×6.65×10(-29)
= 2.22×10^{-6}
acceleration a = F/m = 2.22×10^{-6}/ 5.10×10^{-27}
a= 4.4×10 m/s^2