Answer:
1.) 2.4
2.) 112 lbs
3.) 7.85 inches
4.) 6 lbs
5.) 2 lbs
6.) 67%
Explanation:
Given that
Radius of the wheel R = 1 foot
1 foot = 12 inches.
Radius of the axle r = 5 inches
1.) The mechanical advantage MA is :
MA = R/r = 12/5 = 2.4
2.) How much resistance force can ideally be overcome when an effort of 80 lbs is applied to the wheel of the water valve in problem 1?
MA = Load / effort
Where effort = 80 lbs
Substitute MA and effort into the formula
2.4 = Load / 80
Cross multiply
Load = 2.4 × 80 = 192 lbs
The resistance force to be overcome will be
Force = load - effort
Resistance force = 192 - 80 = 112 lbs
3) What is the linear distance traveled when a 2.5' diameter wheel makes one revolution
One revolution = 2π
Radius = 2.5 /2 = 1.25 inches
Linear distance S = angular distance Ø × radius r
S = Ør
S = 2π × 1.25
S = 7.85 inches
4. ) given that
Wheel radius R = 4
Axle radius r = 1
MA = 4/1 = 4
MA = Load / effort
4 = 24/ effort
Effort = 24/4 = 6 lbs
5.) 6 - 4 = 2lb
6.) Efficiency = MA / VR × 100
Efficiency = 4 / 6 × 100
Efficiency = 67%
Answer:
Composite panel garage doors
Explanation:
Answer:
I know ship has stream line body
Answer:
porosity = 0.07 or 7%
dry bulk density = 3.25g/cm3]
water content =
Explanation:
bulk density = dry Mass / volume of sample
dry mass = 0.490kg = 490g
volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3
density = 490/150.8 = 3.25g/cm3
porosity =
=
= 0.07 or 7%
water content =
= 7%
The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.
Define work. Explain the rate of doing work.
Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.
Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.
Solution Explained:
Given,
Weight = 1000N and distance = 5m
A/Q, the work here is done in lifting then
Work = (weight) × (distance moved)
= 1000 X 5
= 5000Nm or 5000J = 5kJ
Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.
To learn more about work, use the link given
brainly.com/question/25573309
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