Answer:
Force on front axle = 6392.85 N
Force on rear axle = 8616.45 N
Explanation:
As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels
Now we know that
now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle
so we can write torque balance about its center of mass
now from above equation
now we have
now the other force is given as
If you go to a website known as 'Wolfram' and type in the question they should be able to answer it. It's basically a very smart calculator. I'd add a link but I don't want to seem like I'm advertising, and it's against Brainly rules.
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:
The distance of separation is 
Explanation:
The mass of the each ball is 
The negative charge on each ball is 
Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero
Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e
=> 
here k the the coulomb's constant with a value 
So
![0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}](https://tex.z-dn.net/?f=0.01%20%2A%209.8%20%20%3D%20%20%5Cfrac%7B%209%2A10%5E9%20%2A%5B1%2A10%5E%7B-6%7D%20%2A%201%2A10%5E%7B-6%7D%5D%7D%7Bd%7D)
1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle) to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.
angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)
sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N
use formula: sin or cos
i used sin. sin(40) = 4905 / ?
- times '?' on both sides. : sin(40) x '?' = 4905
-divide both sides by sin(40): '?' = 4905/ sin(40)
--> Solve.
