0
Still curious? Ask our experts.
Chat with our AI personalities
Maxine
Rafa
FranAdd your answer:
Earn +20 pts
Work of 1 Joules is done in stretching a spring from its natural length to 12 cm beyond its natural length what is the force in Newtons that holds the spring stretched at the same distance 12 cm?
First, set up the integral W= /int_{0}^{0.12}k*x. Solve the integral on the right side. It should lead to W= (9/1250)*k. Substitute W=1 on the left side and solve for k. k should equal (1250/9) Substitute k=(1250/9) and x=0.12 in the equation F=kx and solve for F. F should equal (50/3).
Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 miles down a straight shoreline from the point nearest the boat She can row 2 mph and can walk 4 mph Where should she land her?
r = distance she rowedw = distance she walkedt = total time to reach villaget = r/2 + w/5r^2 = 2^2 + (6 - w)^2r = (4 + 36 - 12w + w^2)^(1/2)r = (w^2 - 12w + 40)^(1/2)t = (w^2 - 12w + 40)^(1/2)/2 + w/5t' = (1/2)(1/2)(w^2 - 12w + 40)^(-1/2)(2w - 12) + 1/5 = 0(1/2)(1/2)(w^2 - 12w + 40)^(-1/2)(2w - 12) = -1/5(1/2)(w - 6) = -1/5(w^2 - 12w + 40)^(1/2)5(w - 6) = -2(w^2 - 12w + 40)^(1/2)25(w^2 - 12w + 36) = 4(w^2 - 12w + 40)(25 - 4)w^2 + (-25*12 + 4*12)w + 25*36 - 4*40 = 021w^2 - 252w + 740 = 0w = 6.87, 5.13 midiscard 6.87 ( > 6)r = (w^2 - 12w + 40)^(1/2)r = 2.18 miShe should land her boat 5.13 miles away from village
Can anyone find a formula for the described function A rectangle has a perimeter of 20m Express the area of the rectangle as a function of the length of one of its sides?
Suppose the length and width of the rectangle are L and W metres respectively.Then the perimeter, P = 20 m implies that2(L + W ) = 20 => L + W = 10 or W = 10 - L.Then Area = L * W = L * (10 - L) sq metres.
What are domain and range of function graph below?
Type your answer here... C.H(w) > 0
How many calories does it take to raise 1 kg 1 meter?
The work done is transformed into potential energy (in the object lifted) so W = m x g x dz where W = Work m = mass g = acceleration due to gravity dz = change in height => W = 1kg x 1m x 9.81m/s/s => W = 9.81J 1J = 0.239 calories => It takes 2.345 calories to lift 1kg by 1m NOTE These are "real" calories. The calories usually talked about in relation to dieting etc are kilo-calories so to equate this value to the calories in a chocolate bar, divide by 1000
