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From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by ' $m_{best}$ '

and the slope of the worst fit line be represented by ' $m_{worst}$ '

Given that:

$m_{best}$ = 1.35 m/s

$m_{worst}$ = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

$\Delta m = \frac{m_{best}-m_{worst}}{2}$ (1)

Substituting values in eqn (1), we get

$\Delta m = \frac{1.35 - 1.29}{2}$ = 0.03 m/s

8 0

3 years ago

In a sun's life span, if the Hydrogen to Helium ratio at the Red Giant phase is 100% Helium and 0% Hydrogen, Predict the Hydroge

serg [7]

Answer:

3rd bc the atmoce

Explanation:

4 0

2 years ago

A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I

Anuta_ua [19.1K]

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = $\frac{4\times \pi^2\times r^3}{GM}$ where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

8 0

3 years ago

Which of the following objects exerts a gravitational force?

kompoz [17]

the earth exerts a gravitational force

4 0

2 years ago

Read 2 more answers

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago