Answer:
60 km
Explanation:
For an object (or a person, such as in this case) moving at constant speed, the speed is equal to the ratio between the distance travelled and the time taken:
where
v is the speed
d is the distance
t is the time taken
In this case, we have:
v = 120 km/h is the speed
t = 30 min = 0.5 h is the time taken
Therefore, we can rearrange the equation to find the total distance travelled:
Answer:
a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
Explanation:
Given that;
height of the ramp h1 = 0.40 m
foot of the ramp above the floor h2 = 1.50 m
assuming R = 15 mm = 0.015 m
density of steel = 7.8 g/cm³
density of aluminum = 2.7 g/cm³
a) distance that the solid steel sphere sliding down the ramp without friction;
we know that
distance = speed × time
d = vt --------let this be equ 1
according to the law of conservation of energy
mgh₁ = mv²
v² = 2gh₁
v = √(2gh₁)
from the second equation; s = ut + at²
that is; t = √(2h₂/g)
so we substitute for equations into equation 1
d = √(2gh₁) × √(2h₂/g)
d = √(2gh₁) × √(2h₂/g)
d = 2√( h₁h₂ )
we plug in our values
d = 2√( 0.40 × 1.5 )
d = 1.55 m
Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b)
distance that a solid steel sphere rolling down the ramp without slipping;
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c)
distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1 again
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.549√( h₁h₂ )
d = 1.549√( 0.4 × 1.5 )
d = 1.2 m
Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) distance that a solid aluminum sphere rolling down the ramp without slipping.
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m