Does hot water or cold water boil faster is that repeated observations using models are controlled experiments

Read this article discussing the amount of time that people of various ages and education levels report spending on different te

iVinArrow [24]

The article is not found here but surveys are important because they are representative samples of a population.

<h3>What is a survey?</h3>

A survey is a useful tool based on population samples, which is used to make statistical analyses in a given investigation.

Differences in surveys are generally due to small sample sizes, which may lead to errors in the analysis of data.

In conclusion, the article is not found here but surveys are important because they are representative samples of a population.

Learn more about surveys here:

brainly.com/question/13624055

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7 0

1 year ago

A crane lifts a load of 15000N to a height 40m in 50 seconds. Calculate the power of the crane

Nataly [62]

Answer:

P = 12000 W

Explanation:

General data:

F = 15000 N
d = 40 m
t = 50 s
P = ?

Work is force times unit of distance. So in order to calculate the power we must first calculate the work.

Formula:

$\boxed{\bold{W=\frac{F}{d}}}$

Replace and solve

$\boxed{\bold{W=\frac{15000\ N}{40\ m}}}$

$\boxed{\bold{W=600000\ J}}$

once the work is found, we proceed to find the power according to the formula:

$\boxed{\bold{P=\frac{W}{t}}}$

$\boxed{\bold{P=\frac{600000\ J}{50\ s}}}$

$\boxed{\boxed{\bold{P=12000\ W}}}$

The power of the crane is <u>12000 Watts.</u>

Greetings.

6 0

2 years ago

When a 5.0kg cart undergoes a 2.2m/s increase in speed, what is the impulse of the cart

Karolina [17]

Answer:

11.0 kg m/s

Explanation:

The impulse exerted on the cart is equal to its change in momentum:

$I=\Delta p=m\Delta v$

where

m = 5.0 kg is the mass of the cart

$\Delta v=2.2 m/s$ is its change in speed

Substituting numbers into the equation, we find

$I=(5.0kg)(2.2 m/s)=11 kg m/s$

7 0

3 years ago

Read 2 more answers

1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown

denis23 [38]

Answer:

a) 3.37 x $10^{3} kg/m^3$

b) 6.42kg/ $m^{3}$

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x $10^{-4}$ $kg/m^{3}$ . So density of metal = mass of metal / volume of metal = 5 / 14 x $10^{-4}$ $kg/m^{3}$ = 3.37 x $10^{3} kg/m^3$

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/ $m^{3}$

3 0

2 years ago

A large bottle contains 150 L of water, and is open to the atmosphere. If the bottle has a flat bottom with an area of 2 ft, cal

Yuri [45]

Answer:

Total pressure exerted at bottom = 119785.71 N/m^2

Explanation:

given data:

volume of water in bottle = 150 L = 0.35 m^3

Area of bottle = 2 ft^2

density of water = 1000 kg/m

Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water

Pressure due to water P = F/A

F, force exerted by water = mg

m, mass of water = density * volume

= 1000*0.350 = 350 kg

F = 350*9.8 = 3430 N

A = 2 ft^2 = 0.1858 m^2

so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2

Atmospheric pressure

At sea level atmospheric pressure is 101325 Pa

Total pressure exerted at bottom = 18460.71 + 101325 = 119785.71 N/m^2

Total pressure exerted at bottom = 119785.71 N/m^2

6 0

3 years ago