Applied Science, Electrical Principles 1 And Electronics

i. To find the maximum pressure, we can assume that the weight of the box is evenly distributed across the entire bottom surface of the box. In that case, the pressure on the bottom surface would be equal to the weight of the box divided by the area of the bottom surface. The area of the bottom surface is 30 cm * 50 cm = 1500 sq cm, so the maximum pressure would be 70,000 g / 1500 sq cm = 46.67 g/sq cm.

ii. To find the minimum pressure, we can assume that all of the weight of the box is concentrated in a single point at the bottom center of the box. In that case, the pressure on the bottom surface would be equal to the weight of the box divided by the area of a small circle with a radius equal to the distance from the center of the box to any of the corners. The distance from the center of the box to a corner is (85 cm / 2) * sqrt(2) = 59.5 cm, so the radius of the circle would be 59.5 cm / 2 = 29.75 cm. The area of this circle is 29.75 cm * 29.75 cm * pi = 2744.63 sq cm, so the minimum pressure would be 70,000 g / 2744.63 sq cm = 25.47 g/sq cm.

To determine the gas pressure at 95°C, we can use the ideal gas law:

PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas in kelvin.

Since the volume and number of moles of gas are constant, we can rearrange the formula to solve for P:

P = (nRT) / V

We can substitute the given values into the formula to solve for the gas pressure at 95°C:

P = (n * R * (95°C + 273)) / V = (n * R * 368) / V = (304Pa * 368) / V = 112,192 Pa / V

Since the volume of the gas is constant, the pressure of the gas at 95°C will be proportional to the pressure at 23°C. Therefore, the gas pressure at 95°C will be 112,192 Pa / 304 Pa = 368 times the gas pressure at 23°C.

The gas pressure at 95°C is 368 * 304 Pa = 112,192 Pa.

i. To solve this problem, we can use the formulas for mechanical advantage and efficiency:

Mechanical advantage = Output force / Input force

Efficiency = (Output work) / (Input work)

First, we can use the given velocity ratio to calculate the mechanical advantage:

Mechanical advantage = (Output distance) / (Input distance) = 2

Since the mechanical advantage is equal to the velocity ratio, we can conclude that the output distance is twice the input distance.

ii. Next, we can use the given efficiency to calculate the input work:

Input work = Output work / Efficiency = (Output force * Output distance) / Efficiency

Substituting the values given in the problem, we get:

Input work = (400 N * Output distance) / 0.95

Since the output distance is twice the input distance, we can substitute 2 * Input distance for Output distance:

Input work = (400 N * 2 * Input distance) / 0.95

Solving for the input distance, we get:

Input distance = Input work / (400 N * 2 / 0.95) = 509.2 m

The input distance is equal to the input force applied to the inclined plane, so the input force is 509.2 N.

To calculate the effort required, we can use the formula for mechanical advantage:

Effort = Load / Mechanical advantage = 400 N / 2 = 200 N

The effort required to raise the load is 200 N.