Answer:
$364.29
Explanation:
given,
Packing of crates per month (u)= 800
annual carrying cost of 35 percent of the purchase price per crate.
Ordering cost(S) = $ 28
D = 800 x 12 = 9600 crates/year
H = 0.35 P
H = 0.35 x $10
H = $3.50/crate per yr.
Present Total cost
=
= 1400 + 336
= $ 1,736
Total cost at EOQ
=
= 685.86 + 685.85
= $ 1,371.71
the firm save annually in ordering and carrying costs by using the EOQ
= $ 1,736 - $ 1,371.71
= $364.29
Answer:
(a) True
Explanation:
A supernova is a powerful light explosion that occurs in massive stars.
During a supernova, the star releases very large amounts of energy as well as neutrons, which allows elements heavier than iron, such as silver, uranium and lead, to be produced.
Therefore, the correct option is "a" True
Elements more massive than iron (e.g. silver, uranium and lead) are created from supernova events.
The only thing acting as the roller coaster gets to the bottom is gravity thus
acceleration is 9.81
so u use the formula
vf^2 = vi^2 + 2abetaD
thus
distance = vf^2 - vi^2 divided by 2a
so 26^2 - 0^2 divided by 2(9.81m/s)
equals 676 divided by 2(9.81m/s)
= 34.45463812m
then use the appropriate 3 sig digs
so 34.5m
hopes this helps
Answers:
The acceleration due to gravity on the surface of earth is 9.8 ms^(-2).Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon . <br> Hint : T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m))) <br> (T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e)) <br> g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)
Answer:
So, the car should be travelling at a speed of 24.82 m/s
Explanation:
Using Δy = y₂ - y₁ = v₁t + 1/2at², We find the time it takes the car to land on the other side.
Given y₂ = 1.5 m
(the height of the other side above the river), y₁ = 23.0 m (the height of the car's side above the river), v₁ = initial vertical velocity = 0, a = -g = -9.8 m/s²
Δy = v₁t + 1/2at² = 0 - 1/2gt² = -1/2gt²
t = √-(2Δy/g) =√-(2(y₂ - y₁)/g) = √-(2(1.5 - 23.0)/9.8) = √-2(-21.5)/9.8) = √(43/9.8) = √4.388 = 2.095 s ≅ 2.10 s
We now find the initial horizontal velocity u₁ using Δx = x₂ - x₁ = u₁t + 1/2at²
Given Δx = 52.0 m the width of the river, a = 0 (no horizontal acceleration) and t = 2.095 s
Δx = u₁t + 1/2at² = u₁t + 0 = u₁t
u₁ = Δx/t = 52.0 m/2.095 s = 24.82 m/s
So, the car should be travelling at a speed of 24.82 m/s just as it leaves the cliff in order to just clear the river and land safely on the other side.