This is a problem of conservation of momentum
Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s
A) man throws the rock forward
=>
rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man
sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?
Conservation of momentum:
momentum before throw = momentum after throw
46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2
=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s
B) man throws the rock backward
this changes the sign of the velocity, v2 = -14.5 m/s
46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2
v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s
Answer:
h = 9.83 cm
Explanation:
Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water
let's reduce the magnitudes to the SI system
r = 10 cm = 0.10 m
m = 10 g = 0.010 kg
A = 100 cm² = 0.01 m²
the definition of density is
ρ = m / V
the volume of a sphere
V = 
V = 
π 0.1³
V = 4.189 10⁻³ m³
let's calculate the density of the ball
ρ = 
ρ = 2.387 kg / m³
the tabulated density of water is
ρ_water = 997 kg / m³
we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation
B - W = 0
B = W
where B is the thrust that is given by Archimedes' principle
ρ_liquid g V_submerged = m g
V_submerged = m / ρ_liquid
we calculate
V _submerged = 0.10 9.8 / 997
V_submerged = 9.83 10⁻⁴ m³
The volume increassed of the water container
V = A h
h = V / A
let's calculate
h = 9.83 10⁻⁴ / 0.01
h = 0.0983 m
this is equal to h = 9.83 cm
Answer:
A simple machine consisting of an axle to which a wheel is fastened so that torque applied to the wheel winds a rope or chain onto the axle, yielding a mechanical advantage equal to the ratio of the diameter of the wheel to that of the axle.
Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s
It was a man named <span>Johannes Kepler. </span>
