As per kinematics equation we are given that
now we are given that
a = 2.55 m/s^2
now we need to find x
from above equation we have
so it will cover a distance of 93.2 m
Answer:
Incomplete question
This is the complete question
For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter
Explanation:
Given the magnetic field
B=2T
Lenght of rod is 1mm
L=1/1000=0.001m
Diameter of rod=1.5mm
d=1.5/1000=0.0015m
Radius is given as
r=d/2=0.0015/2
r=0.00075m
Area of the circle is πr²
A=π×0.00075²
A=1.77×10^-6m²
Given that the voltage applied is 100mV
V=0.1V
Given that resistive is 0.6 Ωm
We can calculate the resistance of the cylinder by using
R= ρl/A
R=0.6×0.001/1.77×10^-6
R=339.4Ω
Then the current can be calculated, using ohms law
V=iR
i=V/R
i=0.1/339.4
i=2.95×10^-4 A
i=29.5 mA
The force in a magnetic field of a wire is given as
B=μoI/2πR
Where
μo is a constant and its value is
μo=4π×10^-7 Tm/A
Then,
B=4π×10^-7×2.95×10^-4/(2π×0.00075)
B=8.43×10^-8 T
Then, the force is given as
F=iLB
Since B=2T
F=iL(2B)
F=2.95×10^-4×2×8.34×10^-8
F=4.97×10^-11N
Answer:
Taking forces along the plane
F cos θ - M g sin θ -100 = M a net of forces along the plane
F = (M a + M g * .5 + 100) / .866 solving for F
F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N
F = 707 N acting along the plane
Fn = F sin θ + M g cos θ forces acting perpendicular to plane
Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons forces normal to plane
(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)
Answer:
The answer to this is falling all the way through the Earth is impossible, since its core is molten. ... As you approached the center of the earth the pull of gravity would decline and eventually (at the center) cease, but inertia would keep you going.
Explanation:
your welcome
The downward pull of an object due to gravity is the object’s weight.
