<u>Option a.</u>The particle of larger mass has more momentum.
How this is explained?
- Given two particles are of different mass & start from rest .
- They have same net force & distance travelled due to force is also constant.
- We know Kinetic energy
- and momentum p = mv .
- The same gives larger mass a smaller acceleration.
- The body which has larger mass will take a longer time interwal to move through same distance; the impulse given to larger mass is larger, thus larger mass will have a greater final momentum.
- Thus option(a).
What is a momentum?
- In Newtonian mechanics, linear momentum, translational momentum, or simply momentum is the product of the mass and velocity of an object.
- It is a vector quantity, possessing a magnitude and a direction.
- If m is an object's mass and v is its velocity (also a vector quantity), then the object's momentum p is :p=mv.
- In the International System of Units (SI), the unit of measurement of momentum is the kilogram metre per second (kg⋅m/s), which is equivalent to the newton-second.
To know more about momentum, refer:
brainly.com/question/1042017
#SPJ4
A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² -
r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄
B) E = kQ/d²
since the distribution is symmetric spherically
C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 -
r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.
D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a
min).
<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>
1 nanometer =
1 × 10-7 centimeter
Answer:
a) , b)
Explanation:
a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:
The force on the skier is:
b) The equations of equilibrium are the following:
The force on the skier is: