What is the answer to these problems 30 pt

2. Recall the equation for an object's gravitational potential energy (assuming relative to the surface of the earth):

PE = mgh

PE is the potential energy.

m is the object's mass.

g is the acceleration due to earth's gravity.

h is the object's height from the earth's surface.

Given values:

PE = 19600 J

m = 100 kg

g = 9.81 m/s²

Substitute the terms in the equation with the given values and solve for h:

19600 = 100×9.81×h

3. Recall the equation for an object's kinetic energy:

KE = 0.5mv²

KE is the kinetic energy.

m is the object's mass.

v is its velocity.

Given values:

KE = 25 J

m = 2 kg

Substitute the terms in the equation with the given values and solve for v:

25 = 0.5×2v²

Use the equation for an object's potential energy again:

PE = mgh

Given values:

PE = 98 J

m = 2 kg

g = 9.81m/s²

Substitute and solve for h:

98 = 2×9.81×h

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm