The effort distance of a lever should be increased to lift a heavier load because it provides a mechanical advantage, allowing for easier lifting of the load.
The effort distance of a lever should be increased to lift a heavier load because it allows for a mechanical advantage that compensates for the increased weight.
In a lever system, the effort distance is the distance between the point of application of the input force (effort) and the fulcrum, while the load distance is the distance between the point of application of the output force (load) and the fulcrum. The mechanical advantage of a lever is determined by the ratio of the load distance to the effort distance.
By increasing the effort distance, the mechanical advantage of the lever system is increased. This means that for the same input force (effort), a greater output force (load) can be achieved. When dealing with a heavier load, a higher mechanical advantage is required to overcome the increased resistance.
By increasing the effort distance, the lever system can effectively multiply the applied force, making it easier to lift the heavier load. This allows for the redistribution of force and facilitates the efficient use of human effort in various applications, such as in construction, engineering, and even everyday tools like scissors and pliers.
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A hollow aluminum cylinder 18.0 cm deep has an internal capacity of 2.000 L at 15.0°C. It is completely filled with turpentine at 15.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 85.0°C. (The average linear expansion coefficient for aluminum is 24 ✕ 10^−6°C^−1, and the average volume expansion coefficient for turpentine is 9.0 ✕ 10^−4°C−1.) Answer parts a-c.
Answer:
a) Calculate the change in the radius of the cylinder between 15.0°C and 85.0°C.
Given:
Depth of cylinder = 18.0 cm = 0.180 m
Average linear expansion coefficient for aluminum = 24 x 10^-6 /°C
Temperature change = 85 - 15 = 70 °C
Change in radius = (initial radius) x (linear expansion coefficient ) x (temperature change)
= (0.180/π) x (24 x 10^-6) x (70)
=2.16 x 10^-4 m = 0.0216 mm
b) Calculate the change in volume of the turpentine between 15.0°C and 85.0°C.
Given:
Initial volume of turpentine = 2.000 L
Average volume expansion coefficient for turpentine = 9.0 x 10^-4 /°C
Temperature change = 85 - 15 = 70 °C
Change in volume = (initial volume) x (volume expansion coefficient) x (temperature change)
= 2.000 L x (9.0 x 10^-4) x 70
= 0.126 L
c) Will any turpentine overflow? Explain your reasoning.
No turpentine will overflow because the increase in the radius of the cylinder is greater than the increase in the volume of the turpentine.
The cylinder radius increases by 0.0216 mm (part a) while the volume of turpentine increases by only 0.126 L (part b). This indicates the expanded cylinder can accommodate the increased volume of turpentine, so no overflow will occur.
Explanation:
if wrong im sorry
A small ferryboat is 4.50 m wide and 6.70 m long. When a loaded truck pulls onto it, the boat sinks an additional 4.70 cm into the river. What is the weight of the truck? Answer is in kN.
Explanation:
Volume of water displaced = 4.5 m x 6.7 m x .047 m = 1.417 m3
mass of displaced water =
1.417 m3 x 1000kg / m3 = 1417 kg = 13900 N = 13.9 kN
The diagram below shows snapshots of an oscillator at different times . What is the frequency of the oscillation ?
The period of the oscillation, given the different time for the oscillator at different stage is 1.80 s
How do i determine the period of the oscillation?Period of an oscillation is defined as the time taken for the oscillation to make one complete circle or oscillation.
With the above information, we can obtain the period of the oscillation. This is illustrated below:
From the diagram given above, we can see that the oscillation started at t = 0.00s and at displacement of +0.10 m. Also, we can see that the oscillation ended at t = 1.80 s and at displacement of +0.10 m.
Now from the above, we can conclude that the period of the oscillation is 1.80 s since the oscillation started at +0.10 m and took 1.80 s to get back to +0.10 m
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In a water pistol, a piston drives water through a larger tube of radius 1.30 cm into a smaller tube of radius 1.10 mm as in the figure below. Answer parts a-f.
It takes 0.47 seconds for water to travel from the nozzle to the ground when the water pistol is fired horizontally.
What is the time it takes for water to travel from the nozzle to the ground?We will denote the height of the water pistol above the ground as h and the initial velocity of water exiting the nozzle as v2. Assuming negligible air resistance, we will analyze the vertical motion of the water droplets.
The vertical displacement of the water droplets is calculated using equation: h = (1/2) * g * t^2.
Rearranging equation, we solve for time:
t = sqrt(2h / g).
Given data:
Height h = 1.10 m and the acceleration due to gravity g = 9.8 m/s^2, we get:t = sqrt(2 * 1.10 / 9.8)
t = 0.47380354147
t = 0.47 seconds.
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A cube of wood having an edge dimension of 20.9 cm and a density of 655 kg/m3 floats on water. Answer parts a-b.
Answer:
Mass of wood = .209^3 m^3 * 655 kg/m^3 = 5.98 kg
area of wood = ,209^2 = .04368 m^3
.04368 * h * 1000 = 5.98 where h is water displaced
h = .137 m = 13.7 cm
a) distance from wood to water = 20.9 - 13.7 = 7.2 cm
Mass of 7.2 cm of wood (wood out of water)
M = 655 kg /m^3 * .209^2 m^2 * .0072 m = .206 kg
.206 kg of Pb must be added to wood for submersion
2. For electric circuit shown in Figure find currents in each resistor.
The current flowing in the 2Ω and 1Ω is 1.14 A and the current flowing in the 3Ω and 4Ω is 0.286 A.
What is the current flowing in each resistor?The value of the current in each resistor is calculated by applying Kirchoff voltage law as follows;
The total voltage in loop 1 is calculated as;
2 + 4 - I₁R₁ - (I₁ - I₂)R₂ - I₁R₃ = 0
6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0
The current flowing in loop 2 is calculated as;
I = V/R
I₂ = ( 6 V - 4 V ) / (3 + 4)
I₂ = 0.286 A
The value of the current flowing in loop 1 is calculated as;
6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0
6 - 2I₁ - 3(I₁ - 0.286) - 1₁ = 0
6 - 3I₁ - 3₁ + 0.858 = 0
-6I₁ = -6.858
I₁ = 6.858 / 6
I₁ = 1.14 A
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what impact does liberal arts have on ensuring continued innovation
Answer:
Liberal arts education is important for ensuring continued innovation. It helps students think creatively, solve problems, collaborate effectively, and consider ethical factors. By exploring various subjects and adapting to new situations, liberal arts education equips individuals with the skills needed to generate new ideas and drive progress in different fields.
What is indicated by the following displacement time graph ?
Answer:
The object in motion is deaccelerating in the negative direction.
Explanation:
Interpret the given position-time graph.
Observing the graph we can determine the slope is negative and is gradually flattening out. Thus, we can conclude the object in motion is deaccelerating in the negative direction.
Question 1
At one section of a long pipe the velocity of the fluid is 1.6 m/s. At another section of the pipe the diameter is three times greater.
What is the velocity of the fluid at this section?
O 0.533 m/s
○ 4.80 m/s
O Not enough information to tell
O 0.178 m/s
Question 2
Three thermometers are placed in a closed, insulated box and are allowed to reach thermal equilibrium. One is calibrated in
Fahrenheit degrees, one in Celsius degrees, and one in Kelvins. If the Celsius thermometer reads -40 °C the Fahrenheit
thermometer would read -40°F.
True
False
Answer:
Answer 1: The answer is O 0.178 m/s.
Answer 2: True: But in this specific case where the Celsius temperature is -40, the Fahrenheit temperature will also be -40.
So, in short, the answer is:
-40 Celsius is equal to -40 Fahrenheit
A steam turbine receives steam with a velocity of 28 m/s, specific enthalpy 3000 kJ/kg at a rate of 3500 kg per hour. The steam leaves the turbine with a specific enthalpy of 2200 kJ/kg at 180 m/s. Calculate the turbine output, neglecting losses.
A steam turbine receives steam with a velocity of 28 m/s, specific enthalpy 3000kJ/kg at a rate of 3500 kg per hour. The steam leaves the turbine with a specific enthalpy of 2200kJ/kg at 180 m/s then turbine output is 777.76kW.
To get the turbine output, we must first compute the change in specific enthalpy (h) and mass flow rate ().
Assume that the inlet steam velocity (v1) is 28 m/s.
Specific enthalpy at the inlet (h1) = 3000 kJ/kg
()=3500kg/h mass flow rate
2200 kJ/kg outlet specific enthalpy (h2)
v2 (outlet steam velocity) = 180 m/s
To begin, convert the mass flow rate from kg/h to kg/s as follows: =
[tex]3500 kg/h (1 h/3600 s) = 0.9722 kg/s[/tex]
The change in specific enthalpy (h) can then be calculated:
3000kJ/kg-2200kJ/kg=800kJ/kgh=h1-h2
The following formula can be used to compute the turbine output (P):
[tex]P = ṁ * Δh[/tex]
Substituting P=0.9722kg/s*800kJ/kg=777.76kJ/sork W
As a result, ignoring losses, the turbine output is roughly 777.76kW
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Question 13
1.75 pts
How much heat is required to heat 0.44 kg of ice from -20 °C to water at 30°C. Numerical answer is assumed to be given in kJ.
Specific heat of ice is 2090 J/(kg "C), specific heat of water is 4186 J/(kg °C)
Latent heat of Fusion of Water is 3.33 *10^5 J/kg
Latent heat of vaporization of Water is 2.26*10^6 J/kg
The amount of heat required to heat the ice from -20 °C to water at 30°C is 238,612 J.
What is the quantity of heat required?The amount of heat required to heat the ice from -20 °C to water at 30°C is calculated as follows;
Q = Q₁ + Q₂ + Q₃
where;
Q₁ is the heat required to raise the -20⁰c to ice at 0⁰CQ₂ is the heat required to melt the ice at 0⁰CQ₃ is the heat required to raise the liquid at 0⁰C to 30⁰CThe amount of heat required to heat the ice from -20 °C to water at 30°C is calculated as;
Q = (0.44 x 4186 x 20) + (3.33 x 10⁵ x 0.44) + (0.44 x 4186 x 30)
Q = 238,612 J
Thus, the total quantity of heat required to raise the temperature of the ice to the liquid is 238,612 J.
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