Answer:
100
Explanation:
= Resistivity of axon
r = Radius of axon
t = Thickness of the membrane
= Resistivity of the axoplasm
Speed of pulse is given by
So, radius is given by
If radius is increased by a factor of 10 new radius will be
So, The radius will increase by a factor of 100.
Answer:
F = 0.414 N
Explanation:
Given that,
Magnetic flux density,
The length of the wire, l = 24 m
Current, I = 0.48 A
We need to find the force acting on the wire. The formula for the force is given by:
Put all the values,
So, the force acting on the copper wire is equal to 0.414 N.
Answer:
The minimum initial speed of the cannon ball is approximately 99.018 m/s
Explanation:
The parameters of the cannon are;
The maximum horizontal distance reached, R = 3,279 ft. ≈ 999.4392 m
The maximum horizontal range, , is given by the following formula;
Where;
u = The initial speed
g = The acceleration due to gravity ≈ 9.81 m/s²
= 999.4392 m
We get;
u = √(g × )
∴ u = √(9.81 m/s² × 999.4392 m) ≈ 99.018 m/s
The minimum initial speed of the cannon ball (just as it leaves the cannon) that is needed for it to reach this distance, in m/s, u ≈ 99.018 m/s.
In this case, the movement is uniformly delayed (the final
rapidity is less than the initial rapidity), therefore, the value of the
acceleration will be negative.
1. The following equation is used:
a = (Vf-Vo)/ t
a: acceleration (m/s2)
Vf: final rapidity (m/s)
Vo: initial rapidity (m/s)
t: time (s)
2. Substituting the values in the equation:
a = (5 m/s- 27 m/s)/6.87 s
3. The car's acceleration is:
a= -3.20 m/ s<span>^2</span>
Answer:
Explanation:
Given that,
Diameter of pipe
d = 20cm
Then, radius =d/2 =20/2
r = 10cm =0.1m
The speed at the bottom is
Vi = 3m/s
Speed at the top Vf?
At the bottom the cube is at a height of 0m
Then, y1 = 0m
At the top the cube is at a height which is the same as the diameter of the pipe
y2 = 0.2m
Now, let us consider, the energy conservation equation , which is the sum of kinetic energy and gravitational potential energy, given by,
K2 + U2 = K1 + U1
½m•Vf² + m•g•y2 = ½m•Vi² + m•g•y1
Divide all through by m
½•Vf² + g•y2 = ½•Vi² + g•y1
Since y1 = 0
So we have,
½•Vf² + g•y2 = ½•Vi²
½•Vf² = ½•Vi² — g•y2
Multiply through by 2
Vf² = Vi² —2g•y2
Vf = √(Vi²—2g•y2)
g is a constant =9.81m/s2
Vf = √(3²—2×9.81×0.2)
Vf = √(9—0.981)
Vf = √8.019
Vf = 2.83m/s
The speed of the ice cubes at the top of the pipe is 2.83m/s