Answer:
Explanation:
At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation
Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.
Simplifying
v = 
Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get
v = ![[tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B2%5Ctimes%20%281.2%5Ctimes10%5E%7B-5%7D%29%5E2%282182-1.225%29%7D%7B9%5Ctimes%201.8%5Ctimes10%5E%7B-5%7D%7D)
[/tex]
v = 387 x 10⁻⁵ m/s
Terminal velocity = 387 x 10⁻⁵ m/s
Time taken to fall a distance of 100 m
= 
= 2.6 x 10⁴ s.
The magnetic force experienced by the proton is given by
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so
and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
So we have
from which we can find the magnitude of the field:
Answer:
here given is a weight
then force becomes mg
that is F=Mg
=4*9.8
then by using the formula
F=Ma
a=F/M
=4*9.8/9.8
=4
Explanation:
Answer:
b. Friction decreased when he went from pavement to ice and then increased two more times.
Explanation:
Frictional force depends on the normal force of the surface and a friction coefficient.
Since we're talking about the same car, the value of 
will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.
After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.
