What two sysrems of a rabbit's body must be working together to run away?

2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the

frozen [14]

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 = $\frac{1}{2}$ mv²

v² = 2gh₁

v = √(2gh₁)

from the second equation; s = ut + $\frac{1}{2}$ at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

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 = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ 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 = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{3}$ mR²) ω²

v = √( $\frac{6}{5}$ gh₁ )

so we substitute √( $\frac{6}{5}$ gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( $\frac{6}{5}$ 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 = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ 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

8 0

3 years ago

A miner develops cancer of the esophagus. Ten years before, he had been exposed to radiation when he worked for a year in a mine

slamgirl [31]

D.) It is unlikely that a specific cause can be determined, but the treatment would likely be the same in either case.

3 0

3 years ago

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Scientist have measured that the vibrations of an earthquake travel faster through solid rocks in the Earth and slower in the li

GarryVolchara [31]

A. The particles are packed more tightly in materials with more density which causes the vibrations to bounce of the partials more rapidly which makes them go faster

6 0

3 years ago

7) List three (3) automobile safety features currently used to minimize the risk of injury to its passengers. Relate these

r-ruslan [8.4K]

Seatbelt- strapped in egg
air bag- cushion around the egg
brakes- parachute(bag that helps the egg go down slower)

3 0

3 years ago

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle

Alex73 [517]

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

$T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N$

6 0

3 years ago