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Physics questions

A man jumps a fence 2.45 m. If the PEg at the top point of his trajectory was 1.59x10^3 J, what was his mass? a.) 66.2 kg b.) 666 kg c.) 662 kg d.) 26.6 kg
You slide a 5-kg object up a ramp inclined at 25 degrees. How long is the ramp if the PEg of the object is 2.4x10^2 J? A.) 32 m b.) 22 m c.) 12 m
A 100,000 lb freight car is drawn 2500 feet up a long a 1.2% grade at constant speed. Find the work find against gravity. A.) 300,000 J b.) 2,940,000 J c.) 0 N d.) 4,067,400
A 200 N box is pushes up and include 10 long and 3 m high. The average force is 120 N. a.) How much work is done? b.) What is the change in the PE of the box? c.) What is the change in KE how the box? d.) What is the frictional force on the box?
A bullet going 153 m/s crashes through a wood plank. After passing through the plank it's speed is 130 m/s  Another bullet, of the same mass and size traveling at 92 m/s is also fired at the plank. What will the second bullet's speed be after tunneling through? 1.) the resistance of the plank is independent of the speed of the bullet. 2.) the plank does the same amount of work on the two bullets.
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2 Answers

 
Hey Sierra -- 1600J = mgh ... mg= 1600/2.5 ~ 3200/5 ~ 640N ==> a) 66kg mass 
 
240J = mgh ... h= 240/50 ~ 5m high ... sin 25 ~ 25/60 ~ 0.4 = 5m/L => c) 12m up
 
1.2% grade raise car 30ft ... W= mgh ~ 3 million ft-lbs(m/3.3ft)(4.5N/lb) ==> d) 4MJ
 
a) W= F*d = 120(10) = 1200J ... b) PE= mgh = 200(3) = 600J ... c)d) downramp force is 30% of 200N or 60N ... upramp F is 120N => 60N remains to increase KE or to overcome friction -- the split depends on how slippery the ramp is (K sliding) ... if friction is zero, KE increases 600J ... if friction is 60N, KE does not increase -- 120N up = 60wt + 60friction for no net accel up & no speed (KE) change  
 
V1 decrease is 15% ... V1out ~ 5/6 V1in ... KE1out ~ 25/36 KE1in ==> about 30% KE1in lost to "plank work" ... 2nd bullet has 60% of 1st's speed ... KE2in ~ 0.36 KE1in ... deduct 0.30 KE1in for plank => KE2out is 0.06 KE1in ==> V2out ~ sqrt[0.06] V1in ~ 0.25 V1in ~ 153/4 ==> about 38 m/s ... Regards, ma'am :)
a)mgh=1.59 x 10^3
m=(1.59 x10^3)/(9.8*2.45)
m = 66.2Kg
 
ANSWER A
 
b) mgh=2.4x10^2
h = (2.4x10^2)/(5*9.8)
h = 4.89m.
 
sin 25 = 4.89/x ; x = 11. 64m 
 
ANSWER = B
 
 
 
 

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