Answer:
The peak current carried by the axon is 5.85 x 10⁻⁸ A
Explanation:
Given;
distance of the field from the axon, r = 1.3 mm
peak magnetic field strength, B = 9 x 10⁻¹² T
To determine the peak current carried by the axon, apply the following equation;
where;
B is the peak magnetic field
r is the distance of the magnetic field from axon
μ is permeability of free space = 4π x 10⁻⁷
I is the peak current
Re-arrange the equation and solve for "I"
Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A
The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
A) To true. he pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases
Explanation:
Let us propose the solution of this problem before seeing the final statements. The pressure increases with the depth of raposin due to the weight of water that is above the person and also the pressure exerted by the atmosphere on the entire pool, the equation describing this process is
P =
+ ρ g y
Where is the atmospheric pressure, ρ the water density, and 'y' the depth measured from the surface.
Let's examine this equation in we see that the total pressure is directly proportional to the atmospheric pressure and depth
Now we can examine the claims
A) To true. State agreement or with the equation above
B) False. Pressure changes with atmospheric pressure
C) False. It's the opposite
D) False. They are directly proportional
Answer:
Explanation:
Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.
For faster car on the road,
v = 2v
..........(1)
For the slower car on the road,
............(2)
Equation (1) becomes,
So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.
gravitational potential energy is given by formula
here we need to compare the gravitational potential energy of stone 2 with respect to stone 1
so we will say
given that
now we have
