looking for help

[JasonDunn11]

The coal mine you manage has coal cars that are 3 feet wide by 4 fet long ad 1.5 feet deep. If a ton of coal equals 25 cubic feet, how many cars must be filled to complete an order for 15 tons of coal?

I have a trig questions that asks what the RPM of a electric lawn mower blade is if the radius equals 1 and the speed of the blade is 1100 feet per sec. So far I have come up with (want to see if this is correct)

one revolution = 2pie radians(R x Pie)

So I converted feet per second into feet per minute

66000 feet per min by 6.283 (2pie radians) and came up with: 10504 rpm.

Can anyone verify that I did this correct? Need help to solve these 2 question.thanks in advance helping me.

 

[Deleted member 4993]

The coal mine you manage has coal cars that are 3 feet wide by 4 fet long ad 1.5 feet deep. If a ton of coal equals 25 cubic feet, how many cars must be filled to complete an order for 15 tons of coal?

I have a trig questions that asks what the RPM of a electric lawn mower blade is if the radius equals 1 and the speed of the blade is 1100 feet per sec. So far I have come up with (want to see if this is correct)

one revolution = 2pie radians(R x Pie)

So I converted feet per second into feet per minute

66000 feet per min by 6.283 (2pie radians) and came up with: 10504 rpm.

Can anyone verify that I did this correct? Need help to solve these 2 question.thanks in advance helping me.

Prob#1

How many cu.ft of coal goes into each car? (V = width*length*depth cu.ft.)

What is the weight of the full-car of coal? (W = V/25 tons)

# of cars needed = 15/W

Prob #2 (very ill posed problem. Does not define blade-speed location and does not define unit of the length of the blade)

Assuming that the problem is referring to the tip-speed of the blade:

v = r * w

where

v = tip speed (ft/sec),

r = radius of the blade (ft),

w = angular speed (rad/sec)

so

1100 = 1 * w

w = 1100 rad/sec = 66000 rad/min = 66000/(2π) rpm = 10504.23 rpm = ~ 11000 rpm