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2 years ago

What is the weight of a 300kg human on earth

Physics

1 answer:

Alborosie2 years ago

5 0

2940‬N

Explanation:

Given parameter:

Mass of the human = 300kg

Unknown:

Weight of human on earth.

Solution:

Mass is a fundamental quantity that measures the amount of matter contained in a substance.

Weight of a body is the force that is acting on it due to the gravitational pull

Weight = mass x acceleration due to gravity

acceleration due to gravity = 9.8m/s²

Inputting the parameters:

Weight = 300 x 9.8 = 2940‬N

Learn more:

Weight and mass brainly.com/question/5956881

#learnwithBrainly

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As we move from Radiowaves to UV waves in electromagnetic spectrum we know that the wavelength is maximum for radio waves while its frequency is minimum

PART (i)

So as we move from radiowaves to UV region

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3 years ago

Un automóvil viaja a una velocidad constante de 120 km/h, al frenar tarda 10 s en detenerse. a) ¿Cuál es la distancia que recorr

belka [17]

Answer:

166.8 m

30.18 m/s

Explanation:

u = Velocidad inicial del automóvil = 120 km/h = $\dfrac{120}{3.6}=33.33\ \text{m/s}$

v = Velocidad final = 0

s = Desplazamiento

a = Aceleración

t = Tiempo tomado = 10 s

De las ecuaciones cinemáticas de movimiento tenemos

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-33.33}{10}\\\Rightarrow a=-3.33\ \text{m/s}^2$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-33.33^2}{2\times -3.33}\\\Rightarrow s=166.8\ \text{m}$

La distancia que tarda el automóvil en detenerse es 166.8 m

Ahora s = 30 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -3.33\times 30+33.33^2}\\\Rightarrow v=30.18\ \text{m/s}$

La velocidad a la que el otro automóvil sería golpeado es 30.18 m/s.

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2 years ago

Consider the following initial rate data for the decomposition of AB to yield A and B: [AB], mol/L Initial Rate, mol/(L. s) 0.20

Zanzabum

Answer : The order of reaction is 2 (second order) and the rate law is, $R=k[AB]^2$

Explanation :

As the expression of rate is,

$R=k[AB]^n$

where,

R = rate

k = rate constant

[AB] = concentration of AB

n = order of the reaction

First we have to calculate the rate for 0.200 mol/L 'AB' concentration with initial rate 0.00320 mol/L.s

$R_1=k[AB_1]^n$

$0.00320=k(0.200)^n$ .............(1)

Now we have to calculate the rate for 0.400 mol/L 'AB' concentration with initial rate 0.0128 mol/L.s

$R_2=k[AB_2]^n$

$0.0128=k(0.400)^n$ .............(2)

Now dividing the equation 1 by equation 2, we get the order of reaction.

$\frac{0.00320}{0.0128}=\frac{k(0.200)^n}{k(0.400)^n}$

By solving the term, we get the value of 'n'.

n = 2

Thus, the order of reaction = n = 2

The rate law will be,

$R=k[AB]^2$

So we conclude that,

For n = 0, the rate law will be zero order.

For n = 1, the rate law will be first order.

For n = 2, the rate law will be second order.

and so on.....

Hence, the order of reaction is 2 (second order) and the rate law is, $R=k[AB]^2$

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