<span>Answer:
1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use.
Given: m= 1200kg v initial = 95km/hr v final = 0
2nd, focus on the units - in most cases units speak for the concept
the unit of the unknown is kcal, thus its the unit of energy or work
so, W = ?
3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present
since W = delta K.E =delta P.E
W= 0.5m( vf^2 - vi^2) ---> best formula
4th, Substitute the given to the formula
since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J
we express first 95 km/hr to m/s
95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec
W= 0.5(1200kg)[(0^2- (26.39m/sec)^2]
W=600 kg(0 - 696.43m^2/s^2)
W=600kg(-696.43m^2/s^2)
W=417859.3Nm or 417859.3 J
W = 417859.3 J x 1 cal /4.19 J
W = 99,727.7 cal or 99.728 kcal</span>
R(parallel) = product/ sum
50×30/50+30
1500/80
18,75 ohms
Answer:
2.63 %.
Explanation:
Given that,
The calculated value of the specific heat of water is 4.29 J/g.C
Original value of specific heat of water is 4.18 J/g.C.
We need to find the student's percent error. The percentage error in any quantity is given by :
So, the student's percent error is 2.63 %.
KE = 1/2 x 80 x 60^2
KE = 144000
Answer:
3 hours
Explanation:
180 divided by 60 (mph means miles per hours by the way)
