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1 year ago

What is the weight of an object with a mass of 19 kg?

Physics

2 answers:

Natalija [7]1 year ago

7 0

Answer:

Every 2.2 kg is 1 pound. So mulitply 19 * 2.2. It's gonna be equal to 41.8

Explanation:

marysya [2.9K]1 year ago

7 0

Statement:

Mass of an object is 19 Kg.

To find out:

The weight of the object.

Solution:

Mass of the object = 19 Kg

Acceleration due to gravity = $\sf9.8 {m \: {s}^{ - 2} }$

We know, weight = mass × acceleration

Therefore, weight of the object

= (19 × 9.8) N

= 186.2 N

Answer:

The weight of the object is 186.2 N.

Hope you could understand.

If you have any query, feel free to ask.

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

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2 years ago

Which is more useful for storing thermal energy, the block of lead or the water in the calorimeter?

leonid [27]

I believe it is the block of lead

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3 years ago

Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a

FromTheMoon [43]

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

$d_1 = 1* t$

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

$d_2 = 2*(t - 120)$

now it is given that Blaise wins by 10 m distance

$d_1 - d_2 = 10$

$1* t - 2*(t - 120) = 10$

$t - 2t + 240 = 10$

$t = 230 s$

now the distance moved by Blaise is given by

$d_1 = 1*230 = 230 m$

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2 years ago

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dmitriy555 [2]

The correct answer is C; Bicycles.

Further Explanation:

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Bicycle riders are to follow the same laws and laws specifically for the riders or they can face fines and tickets.

Learn more about bicycle laws at brainly.com/question/8934107

#LearnwithBrainly

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Greeley [361]

I think it’s 8 hours. I’m sorry if I’m wrong.
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2 years ago