Answer:
The ladder is moving at the rate of 0.65 ft/s
Explanation:
A 16-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second. We need to find the rate at which the top of the ladder moving down when the foot of the ladder is 5 feet from the wall.
The attached figure shows whole description such that,
.........(1)
We need to find, 
at x = 5 ft
Differentiating equation (1) wrt t as :
Since, 
At x = 5 ft,
So, the ladder is moving down at the rate of 0.65 ft/s. Hence, this is the required solution.
Answer:
The sequence is A,B,H,B,F
Explanation:
- The Standard International unit is Kilogram (kg) and the mass of a body can also be expressed in gram (g).
- Heat is a form of energy and the unit for energy is joule (J), thus the unit of heat is also joule (J).
- Density is mass per unit volume where the unit of mass is gram (g) and the unit of volume can be taken as milli-liter (mL). Thus g/mL is the unit of density.
- The unit of energy is joule (J).
- Molarity is number of solute in mol dissolved in 1 liter of solution. Thus mol/L is the the unit of molarity.
The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.
The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N
The other free body diagram is around the joint of the three strings.
In this case, you can do the horizontal forces equilibrium equation as:
T1* cos(60) - T2*cos(40) = 0
And the vertical forces equilibrium equation:
Ti sin(60) + T2 sin(40) = T3 = 325 N
Then you have two equations with two unknown variables, T1 and T2
0.5 T1 - 0.766 T2 = 0
0.866 T1 + 0.643T2 = 325
When you solve it you get, T1 = 252.8 N and T2 = 165 N
Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N
Answer:
20.42 N/m
Explanation:
From hook's law,
F = ke ......................... Equation 1
Where F = Force applied to the spring., k = spring constant, e = extension.
Make k the subject of the equation,
k = F/e ................. Equation 2
Note: The force on the spring is equal to the weight of the mass hung on it.
F = W = mg.
k = mg/e................ Equation 3
Given: m = 250 g = 0.25 kg, e = 37-25 = 12 cm = 0.12 m.
Constant: g = 9.8 m/s²
Substitute into equation 3
k = (0.25×9.8)/0.12
k = 20.42 N/m.
Hence the spring constant = 20.42 N/m
