Answer:
166.67 N
Explanation:
Given:
Mass flow rate = 50 kg/min = 50 kg / 60 seconds = 0.833 kg/s
Initial velocity = 200 m/s
after striking the normal board the water will flow in the normal direction, thus the final velocity in the direction of the initial flow will be zero 
therefore,
Force = (change in momentum)
or
Force = Initial momentum - final momentum
or
Force = 0.833 × 200 - 0.833 × 0
or
Force = 166.67 N
 
        
             
        
        
        
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:
 v=82 m/s
s=116m
Explanation:
a=20t

using condition given at t=0

c=-8
now equation becomes 
v=10t²-8
v at t= 3s  v=82 m/s
again 


now using condition given s=50 at t=0
b=50
now equation becomes

calculating s at t=3s
s=116m
 
        
             
        
        
        
Answer:
For any string, we use 
Explanation:
The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2. 
Here are the cases:
- Consider any string a i b j c k in the language. If i = 1 or i > 2, we take  and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language. and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
- For i = 2, we can take    and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language. 
- Finally, for the case i = 0, we take  , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well. , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.
 
        
             
        
        
        
Answer:
B. To accurately measure spark advance, use a timing light that incorporates an
ignition advance meter. The spark advance cannot be determined by listening to the way the engine sounds.