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1) A sphere has a radius 1.500m Find its volume in cm^3. Find its suface area in m^2 and in mm^2?

2) A truncated right circular cone has dimensions:
upper base diameter =1.50cm
lower base diameter = 4.00cm
height = 6.00cm
Find its volume in cm^3, m^3, mm^3

3) a person walks 5km in 1,5hr. What is the speed in mi/hr, m/s, and ft/s?

4) A car can accelerate from rest to 60 mi/hr in 7 s. Express this acceleration in the following units(assume constant acceleration)
ft/s^2
km/hr/s
m/s^2

5)A turntable accelerates from rest to 33 rpm in 0.6s. Express this acceleration in rev/s^2 and rad/s^2.

6)A spbway train starts from rest at a station, accelerates uniformly at a rate of 2.0 m/s^2 for 15 s, cruises at constant speed fro 30 s, and then decelerates uniformly coming to rest at the next station after a total trip time of 54 s. How far apart are the stations? Express your answers in m, km and miles.

7)An electrically powered grindstone wheel rotates at 2,000 rpm for 50 s and the power is then switched off. The wheel comes to rest 40.0 s later. Calculate: The angular deceleration in rad/s ^2 and rev/min/s.
Determine the total angular distance traveled over the 90 s time interval, expressed in both radians and revolutions?

https://brainmass.com/physics/units/express-physical-quantities-different-units-measurement-4222

## SOLUTION This solution is FREE courtesy of BrainMass!

1) A sphere has a radius 1.500m Find its volume in cm^3. Find its surface area in m^2 and in mm^2?

r= 1.5 metres
= 150 cm
= 1500 mm
Volume=(4/3)pi r 3= 14137167 cm3
Surface area=4pi r 2= 28.2743 m2
= 28274333.88 mm2

2) A truncated right circular cone has dimensions:
upper base diameter =1.50cm
lower base diameter = 4.00cm
height = 6.00cm
Find its volume in cm^3, m^3, mm^3

To find the volume of the truncated cone
Volume=(1/3)pi r1 2 (h1+h2)-(1/3)pi r2 2 h2=

r1= lower base radius 2 cm
or 0.02 m
or 20 mm

r2= upper base radius 0.75 cm
or 0.0075 m
or 7.5 mm

h1= height of truncated cone 6 cm
or 0.06 m
or 60 mm

h2= height by which the cone has been truncated

h2/(h1+h2)=r2/r1
or h2/(6+h2)=0.75/2= 0.375
or h2= 3.6 cm
or 0.036 m
or 36 mm

Putting these values into the equation
Volume= 38.09181092 cm3
0.000038092 m3
38091.81 mm2

Velocity-acceleration
3) a person walks 5km in 1.5hr. What is the speed in mi/hr, m/s, and ft/s?

1 Km= 1000 metres

1 mile= 1760 yards
= 5280 feet
= 1609.34 metres

1 Km= 0.6214 miles
= 3281 feet

Speed=Distance/time=

Distance= 5 Km Time= 1.5 hrs = 5400 seconds
= 3.107 miles
= 16405 feet

Speed= 2.07 mi/hr
= 0.93 m/s
= 3.04 ft/s
4) A car can accelerate from rest to 60 mi/hr in 7 s. Express this acceleration in the following units(assume constant acceleration)
ft/s^2
km/hr/s
m/s^2

acceleration= change in velocity/change in time
change in velocity= 60-0= 60 mi/hr
= 96.56 Km/hr 1 mile= 1.6093 Km
= 26.82 m/s 1 mile= 1609.34 m 1 hr= 3600 s
= 88 ft/s 1 mile= 5280 ft
change in time= 7 s

acceleration= 12.57 ft/s2
= 13.79 Km/hr/s
= 3.83 m/s2

5)A turntable accelerates from rest to 33 rpm in 0.6s. Express this acceleration in rev/s^2 and rad/s^2.

change in angular velocity ,w = 33 revolutions

change in time= 0.6 s

acceleration= 55.00 rev/s2

6)A subway train starts from rest at a station, accelerates uniformly at a rate of 2.0 m/s^2 for 15 s,
cruises at constant speed from 30 s, and then decelerates uniformly coming to rest
at the next station after a total trip time of 54 s. How far apart are the stations? Express your answers in m, km and mi.

s= ut+ 1/2 a t 2
s= distance travelled
u= initial velocity
t= time
a= acceleration
There are three phases
1)When the train is accelerating
u= 0
a= 2 m/s2
t= 15 s
s= ut+ 1/2 a t 2= 225 metres

2)When the train is moving at constant speed for 30 seconds
After the train has accelerated for 15 seconds the velocity is
v=u+at= 30 m/s

s=vt= 900 metres

3)When the train is decelerating

u= 30 m/s
v= 0 m/s
t= 9 s =54-(15+30); 15 seconds for part 1 and 30 seconds for part 2

v=u+at
a=(v-u)/t= -3.33 m/s2
s= ut+ 1/2 a t 2= 135.14 metres

Total distance= distance travelled in parts 1,2,3= 1260.14 m
= 1.26014 Km
= 0.783050996 mi

7)An electrically powered grindstone wheel rotates at 2,000 rpm for 50 s and the power is then switched off.
the wheel comes to rest 40.0 s later. Calculate: The angular deceleration in rad/s ^2 and rev/min/s.

Determine the total angular distance traveled over the 90 s time interval, expressed in both radians and revolutions?

change in angular velocity= 2000 rpm

change in time= 40 s

angular decelarion=change in angular velocity/change in time=
= 50 rev/m/s = 0.833333333 rev/s^2

During first phase for 50 seconds
angular distance=w t= 1666.67 revolutions

During second phase for 40 seconds
angular distance=w t-1/2*angular deceleration*t^2
= 666.6666667 revolutions

Total= 2333.33 revolutions