Answer:
25.6 V
Explanation:
The kinetic energy of electron associated with its potential difference is given by eV which is equal to the 1/2 mv^2.
m = 9.1 x 10^-31 kg, v = 3 x 10^6 m/s, e = 1.6 x 10^-19 C
eV = 1/2 m v^2
V = mv^2 / 2 e
V = (9.1 x 10^-31) x (3 x 10^6)^2 / (2 x 1.6 x 10^-19)
V = 25.6 V
Answer and Explanation:
In classical physics we study about macroscopic objects and in modern physics we study about microscopic objects but under some circumstances classical physics is a approximation to modern physics such as
- The object must be microscopic but they should be large enough so that we can see that through a microscope
- The speed of the object should be very less it should be about 1 % of the speed of the light
- They should be involved with very week gravitational field
Here in the given situation two forces are acting on the dummy from two perpendicular directions
Since force is a vector quantity so we need to add two or more forces by vector addition method
Now here we know that two forces are perpendicular to each other
So we will use Pythagoras theorem

given that


now plug in all in the above equation


So the sensor will report net force as 4501.9 N
The ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.
<h3>What is the
rotational kinetic energy of a sphere?</h3>
The rotational kinetic energy of the sphere is directly proportional to the square of angular acceleration.

Where,
= rotational inertia of sphere = 2/5MR²
Where
is the mass of the sphere and
is the radius of the sphere,
ω = angular acceleration of sphere = V/R
Where
is the speed of the sphere.
So,
K = 1/2 × 2/5MR² × V²/R²
K = MV²/5
The translational kinetic energy of the sphere,

The total kinetic energy of the sphere,

So,

Thus the Ratio of rotational kinetic energy to total kinetic energy,

Therefore, the ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.
Learn more about rotational kinetic energy:
brainly.com/question/14611300
The question is incomplete. The complete question is :
The pressure difference, Δp, ac
ross a partial blockage in an artery (called a stenosis) is approximated by the equation :

Where V is the blood velocity, μ the blood viscosity {FT/L2}, ρ the blood density {M/L3}, D the artery diameter,
the area of the unobstructed artery, and A1 the area of the stenosis. Determine the dimensions of the constants
and
. Would this equation be valid in any system of units?
Solution :
From the dimension homogeneity, we require :

Here, x means dimension of x. i.e.
![$[ML^{-1}T^{-2}]=\frac{[K_v][ML^{-1}T^{-1}][LT^{-1}]}{[L]}+[K_u][1][ML^{-3}][L^2T^{-2}]$](https://tex.z-dn.net/?f=%24%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D%3D%5Cfrac%7B%5BK_v%5D%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D%5BLT%5E%7B-1%7D%5D%7D%7B%5BL%5D%7D%2B%5BK_u%5D%5B1%5D%5BML%5E%7B-3%7D%5D%5BL%5E2T%5E%7B-2%7D%5D%24)
![$=[K_v][ML^{-1}T^{-2}]+[K_u][ML^{-1}T^{-2}]$](https://tex.z-dn.net/?f=%24%3D%5BK_v%5D%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D%2B%5BK_u%5D%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D%24)
So,
dimensionless
So,
and
are dimensionless constants.
This equation will be working in any system of units. The constants
and
will be different for different system of units.