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3 years ago

7

On the moon what would be the force of gravity acting on an object that has a mass of 7 kg?

1 answer:

Rudiy273 years ago

4 0

Answer:

0

Explanation:

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Sever21 [200]

xin lỗi không có sơ đồ vui lòng cho biết sơ đồ

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3 years ago

A rope is used to pull a 5.07 kg block at constant speed 2.36 m along a horizontal floor. the force on the block from the rope i

mezya [45]

VThe historical method includes what steps?

The historical method includes what steps?

3 0

3 years ago

The rate at which a metal alloy oxidizes in an oxygen-containing atmosphere is a typical example of the practical utility of the

Ray Of Light [21]

Answer:

The activation energy is $Q = 328.31 \ K J/mol$

Explanation:

From the question we are told that

The rate constant is k

at the temperature $T_1 = 300 = 300 + 273 = 573 \ K$

The value of k is $k_1 = 1.05 *10^{-8} \ kg /m^4 \cdot s$

at temperature $T_2 = 400 ^oC = 400 + 273 = 673 \ K$

The value of k is $k_2 = 2.95 *10^{-4} \ kg /m^4 \cdot s$

The rate constant is mathematically represented as

$k = Ce^{- \frac{Q}{RT} }$

Where Q is the activation energy

R is the ideal gas constant with a value of $R = 8.314 \ J /mol \cdot K$

C is a constant

T is the temperature

For the first rate constant

$k_1 = Ce ^{-\frac{Q}{RT_1} }$

For the second rate constant

$k_2 = Ce ^{-\frac{Q}{RT_2} }$

Now the ratio between the two given rate constant is

$\frac{k_1 }{k_2} = e^{(\frac{Q}{R} [\frac{1}{\frac{T_2 - 1}{T_1} } ] )}$

=> $ln [\frac{k_1}{k_2} ] = \frac{Q}{R} * [\frac{1}{\frac{T_2 -1}{T_1} } ]$

substituting values

$ln [\frac{1.05 *10^{-8}}{2.95 *10^{-4}} ] = \frac{Q}{8.314} * [\frac{1}{\frac{673 -1}{573} } ]$

=> $Q = 328.31 \ K J/mol$

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3 years ago

Think critically why would a field guide have common names as well as scientific names?!

Ede4ka [16]

To be able to classify which species is which

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3 years ago

What is it's speed at this time? A rocket blasts off and moves straight upward from the launch pad with constant acceleration. A

zhenek [66]

Answer: 31.85 m/s

Explanation:

Since we are talking about constant acceleration, we can use the following equation to find the speed $s$ of the rocket:

$s=\frac{d}{t}$

Where:

$d=86 m$ is the distance the rocket has traveled

$t=2.7 s$ is the time in which the rocket has traveled the distance $d$

Solving the equation with the given data:

$s=\frac{86 m}{2.7 s}$

$s=31.85 m/s$ This is the rocket's speed

6 0

3 years ago