All categories

3 years ago

If a body travels half its total path in the last 1.50 s of its fall from rest, find the total time of its fall (in seconds).

Physics

1 answer:

svetoff [14.1K]3 years ago

6 0

Answer:

time to fall is 3.914 seconds

Explanation:

given data

half distance time = 1.50 s

to find out

find the total time of its fall

solution

we consider here s is total distance

so equation of motion for distance

s = ut + 0.5 × at² .........1

here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time

so for last 1.5 sec distance is 0.5 of its distance so equation will be

0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)² ........................1

and

velocity will be

v = u + at

so velocity v = 0+ 9.8(t-1.5) ..................2

so first we find time

0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5) + 0.5 ( 9.8)

solve and we get t

t = 3.37 s

so time to fall is 3.914 seconds

You might be interested in

While Newton's Second Law often deals with formulas and numerical calculations, there exist one very important nonnumerical conv

zheka24 [161]

Option B is the correct answer.

MKS system gives the following units:

Distance ----- meters

Mass ----- Kilograms

Time ----- seconds

meter is basic unit for length measurement. smaller units are centimeter, millimeter, micrometer, bigger units are kilometer and so on.

kilogram is the basic unit for mass. smaller unit is gram.

second is the basic unit for time. Greater units are minutes, hours, smallest unit are micro second and so on.

8 0

3 years ago

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

A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleratio

8090 [49]

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

$T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2$

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

4 0

3 years ago

An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act

seropon [69]

Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector $\hat{i}$ .
The positive y-axis = the northern direction, with unit vector $\hat{j}$ .

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
$\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}$

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
$\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})$

The plane's actual velocity is the vector sum of the two velocities. It is
$\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}$

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.

8 0

3 years ago

Which figure represents the electric field lines of two positive point charges?

Naddik [55]

Electric field lines of positive charges are always radially outwards

So here all field lines must have tendency to move out radially as well as two field lines never intersects with each other

So here field lines also do not intersects with each other

So overall the figure C is perfectly representing the field lines of two positive charges as these lines are not intersecting and also originating from two positive charges.

So answer must be

<u><em>FIGURE C</em></u>

7 0

3 years ago