suppose that i suspect that running my ceiling fan stirs up dust and causes my table to become dusty. how could i use the method of concomitant variations to confirm this?

Answers

Answer 1

Answer:

The use the method of concomitant variations we need to identify the variables, establish a baseline, introduce variation and compare results.

To use the method of concomitant variations to confirm whether running your ceiling fan stirs up dust and causes your table to become dusty, follow these steps:

1. Identify the variables: In this case, the independent variable is the operation of the ceiling fan (on or off), and the dependent variable is the dust accumulation on your table.

2. Establish a baseline: Observe the dust accumulation on your table when the ceiling fan is off for a specific period, let's say 24 hours. Document the amount of dust on the table.

3. Introduce variation: Turn on the ceiling fan and observe the dust accumulation on your table for the same period (24 hours). Document the amount of dust on the table.

4. Compare results: Compare the dust levels on your table when the ceiling fan was off versus when it was on. If there is a concomitant (simultaneous) increase in dust accumulation when the fan is on, this suggests a relationship between the operation of the ceiling fan and the dust on your table.

5. Repeat the process: To strengthen your evidence, perform the same experiment multiple times, alternating between turning the fan on and off. If the pattern of increased dust accumulation consistently occurs when the fan is on, this further supports the hypothesis that the ceiling fan stirs up dust and causes your table to become dusty.

To learn more about Concomitant variations, visit:

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State the law of conservation of energy

Answers

Answer:

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. ... For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes.

When the body is in water, how many times faster does it lose heat than when it is in still air? A) 2 B) 10 C) 100 D) 25

Answers

Answer:D)25

Explanation:

I got it right

In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sectional area of 0.075 m2 and a large piston with a cross-sectional area of 0.237 m2 . An engine weighing 3400 N rests on the large piston. What force must be applied to the small piston in order to lift the engine? Answer in units of N.

Answers

Answer:

$F_s=1075.9493\ N$

Explanation:

Given:

area of piston on the smaller side of hydraulic lift, $a_s=0.075\ m^2$

area of piston on the larger side of hydraulic lift, $a_l=0.237\ m^2$

Weight of the engine on the larger side, $W_l=3400\ N$

Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.

$P_s=P_l$

$(F_s)/(a_s)=(W_l)/(a_l)$

$(F_s)/(0.075) =(3400)/(0.237)$

$F_s=1075.9493\ N$ is the required effort force.

Answer:

F = 1076 N

Explanation:

given,

small piston area, a = 0.075 m²

large piston area, A = 0.237 m²

weight on the large piston, W = 3400 N

force applied on the second piston, F = ?

using pascal law for the force calculation

$(F)/(W)=(a)/(A)$

$(F)/(3400)=(0.075)/(0.237)$

F = 0.3165 x 3400

F = 1076 N

The force applied to the small piston in order to lift the engine is equal to 1076 N.

You and a friend ride bicycles to school. Both of you start at the same instant from your house, you riding at 13 m/s and your friend riding at 18 m/s . During the trip your friend has a flat tire that takes him 13 min to fix. He then continues the trip at the same speed of 18 m/s . If the distance to school is 13 km , which of you gets to school first? By comparing the total journey time for you and your friend, this question can be answered.

Answers

V1 = 13 m/s
V2 = 18 m/s
L = 13 km = 13000 m
t0 = 13 min = 780 s

t1 = L/V1 = 13000/13 = 1000 seconds = 16.67 min
t2 = t0 + L/V2 = 780 + 13000/18 = 1500 s = 25 min
You will get to school first

Adding an organism to a food chain does not affect the other members of the food chain, but removing an organism does.

Answers

This is false. Say you have a food chain where a monkey is the top predator. Then you introduce something that eats what the monkey does. The monkey populations will plummet and that means that the things they eat will start to get more numbers. Adding and subtracting animals throws everything out of order.

Removing any level from an ecosystem disrupts a delicate balance that may have ;and a diverse, with many levels and intricate relationships between organisms. More predators kill more prey, which, along with food scarcity, decreases the an attempt to control the population, but it had no effect on habitat degradation.

The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23 to 6°C at an average rate of 10 kg/h. If the COP of this refrigerator is 3.1, the required power input to this refrigerator is (a) 197W
(b) 612W
(c) 64W
(d) 109W
(e) 403W

Answers

Answer:

The correct option is;

Explanation:

Here we have the Coefficient Of Performance, COP given by

$COP = (Q_(cold))/(W) = 3.1$

The heat change from 23° to 6°C for a mass of 10 kg/h which is equivalent to 10/(60×60) kg/s or 2.78 g/s we have

$Q_(cold)$ = m·c·ΔT = 2.78 × 4.18 × (23 - 6) = 197.39 J

Therefore, plugging in the value for $Q_(cold)$ in the COP equation we get;

$COP = (197.39 )/(W) = 3.1$ which gives

$W = (197.39 )/(3.1) = 63.674 \ J$

Since we were working with mass flow rate then the power input is the same as the work done per second and the power input to the refrigerator = 63.674 J/s ≈ 64 W.

The power input to the refrigerator is approximately 64 W.

Answer:

Win = 64 W ... Option C

Explanation:

Given:-

- The water is cooled in the refrigerator with delta temperature, ΔT=(23 - 6 )

- The flow rate of the refrigerated water is flow ( m ) = 10 kg/h

- The COP of the refrigerator is = 3.1:

Find:-

the required power input to this refrigerator is

Solution:-

- The COP - The coefficient of performance of a refrigerator is a quantity that defines the efficiency of the system. The COP is given as:

COP = QL / Win

Where,

QL : The rate of heat loss

Win : The input power required

- The rate of heat loss can be determined from first law of thermodynamics.

Qin - Wout = flow (m)*c*ΔT

Where,

Qin = - QL ... Heat lost.

c : The heat capacity of water = 4,200 J / kg°C

- There is no work being done on the system so, Wout = 0

-QL = flow (m)*c*ΔT

-QL = ( 10 / 3600 )*4200*( 6 - 23 )

QL = 198.33 W

- The required power input ( Qin ) would be:

Win = QL / COP

Win = 198.33 / 3.1

= 63.97 W ≈ 64 W

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