By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span>
=>v¹ = 2.25 m/s (East) </span>
Answer:
Figure attached
We can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.
Explanation:
We can use the following R code to obtain the data for wdiff:
source("http://www.openintro.org/stat/data/cdc.R") #obtain the info
nrow(cdc) # number of elements
names(cdc) # obtain the name for the variable
[1] "genhlth" "exerany" "hlthplan" "smoke100" "height" "weight" "wtdesire" "age"
[9] "gender"
wdiff represent the difference between desired weight (wtdesire) and current weight (weight) and we can obtain the data with the following code:
wdiff <- (cdc$weight-cdc$wtdesire)
And now we can create the histogram with this code
hist(wdiff,xlim =c(-100,150))
> mean(wdiff)
[1] 14.5891
> median(wdiff)
[1] 10
And the result is on the figure attached.
And we can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.
Answer:
1.056 x 10⁷ lb-ft
Explanation:
v = Speed of the bike = 20 mph
t = time of travel = 2 h
d = distance traveled by cyclist
Distance traveled by cyclist is given as
d = v t
d = (20) (2)
d = 40 miles
We know that, 1 mile = 5280 ft
d = 40 (5280) ft
d = 211200 ft
F = force applied by cyclist = 50 lb
W = work done by cyclist
Work done by cyclist is given as
W = F d
W = (50) (211200)
W = 1.056 x 10⁷ lb-ft
Answer:
amount of energy = 4730.4 kWh/yr
amount of money = 520.34 per year
payback period = 0.188 year
Explanation:
given data
light fixtures = 6
lamp = 4
power = 60 W
average use = 3 h a day
price of electricity = $0.11/kWh
to find out
the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66
solution
we find energy saving by difference in time the light were
ΔE = no of fixture × number of lamp × power of each lamp × Δt
ΔE is amount of energy save and Δt is time difference
so
ΔE = 6 × 4 × 365 ( 12 - 9 )
ΔE = 4730.4 kWh/yr
and
money saving find out by energy saving and unit cost that i s
ΔM = ΔE × Munit
ΔM = 4730.4 × 0.11
ΔM = 520.34 per year
and
payback period is calculate as
payback period = 
payback period = 
payback period = 0.188 year