a) 32.3 N
The force of gravity (also called weight) on an object is given by
W = mg
where
m is the mass of the object
g is the acceleration of gravity
For the ball in the problem,
m = 3.3 kg
g = 9.8 m/s^2
Substituting, we find the force of gravity on the ball:
b) 48.3 N
The force applied
The ball is kicked with this force, so we can assume that the kick is horizontal.
This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:
And substituting
W = 32.3 N
Fapp = 36 N
We find
c)
The ball's acceleration can be found by using Newton's second law, which states that
F = ma
where
F is the net force on an object
m is its mass
a is its acceleration
For the ball in this problem,
m = 3.3 kg
F = 48.3 N
Solving the equation for a, we find
To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,
Here,
v = Velocity
= Linear density (Mass per unit length)
T = Tension
Rearranging to find the Period we have that
As we know that speed is equivalent to displacement in a unit of time, we will have to
Therefore the tension is 5.54N
Answer:
160N
Explanation: When 80kg mass is one group . It's reaction force acting on a ground.
Weight of the object = 80*10
= 800 N
Here we are given cofficient of static friction its 0.2. It should be smaller than 1
Friction force = Reaction * Friction Cofficient
Reaction = 800N ( Considering Vertical Equilibrium )
F = 800* 0.2
F = 160N
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>
Answer:
Explanation:
Relative to an origin at the bottom of the hill,
PE = mgh = 10(9.8)(15) = 1470 J