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:
(a) water height =408.66 in.
(b) mercury height=30.04 in.
Explanation:
Given: P=14.769 psi ( 1 psi= 6894.76 
)
we know that 
where 
h=height.
Given that P=14.769 psi ⇒P= 101828.6 7
(a) 
⇒101828.67=
=10.38 m
So water barometer will read 408.66 in. (1 m=39.37 in)
(b) 
=13600
So 101828.67=
=0.763 m
So mercury barometer will read 30.04 in.
Answer:
See explaination
Explanation:
int RED=10; int BLUE=11; int GREEN=12; int BUTTON1=8; int BUTTON2=9; void setup() { pinMode(RED, OUTPUT); pinMode(BLUE, OUTPUT); pinMode(GREEN, OUTPUT); pinMode(BUTTON1, INPUT); pinMode(BUTTON2, OUTPUT); } void loop() { int BTN1_STATE=digitalRead(BUTTON1); int BTN2_STATE=digitalRead(BUTTON2); if(BTN1_STATE==HIGH) { digitalWrite(BLUE, HIGH); delay(1000); // Wait for 1 second digitalWrite(BLUE, LOW); } if(BTN2_STATE==HIGH) { digitalWrite(RED, HIGH); delay(4000); // Wait for 4 seconds digitalWrite(RED, LOW); } if(BTN1_STATE==HIGH && BTN2_STATE==HIGH) { digitalWrite(GREEN, HIGH); delay(2000); // Wait for 2 second digitalWrite(GREEN, LOW); } }
