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1 year ago

Find the work done on a 50 Kg student by the elevator in 2 seconds, if the elevator is :

1 answer:

5 0

The work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻² would be

The total amount of energy transferred when a force is applied to move an object through some distance.

The work done is the multiplication of applied force with displacement.

Work Done = Force × Displacement

As given in the problem we have to find the work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻²,

Lets us first calculate the displacement of the elevator in 2 seconds if it is accelerating upwards from rest at a rate of 2 ms⁻².

s = ut + 0.5at²

= 0 + 0.5×2×2²

= 4 meters

The work done on the elevator = mgh

=50×9.81×4

=1962 joules

Thus, the work done on the elevator would be 1962 joules.

To learn more about work done, refer to the link;

brainly.com/question/13662169

#SPJ1

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What social science, which deals with the behavior and thinking of organisms, differs from sociology primarily in that it focuse

Svetlanka [38]

Answer : The correct answer is- Psychology.

As per the American Psychological Association, psychology can be described as the scientific study of mind and behavior of a particular individual.

There are primarily four goals of psychology that are to describe, to explain, to predict, and to control the mental processes as well behavior of a person.

On the contrary, sociology focuses on behaviour of group of individuals. It is the study of structure, function as well as development of society.

Thus, psychology is the right answer to the question.

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4 years ago

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Which of the following positions would Earth be in if it were winter in the Northern Hemisphere?

abruzzese [7]

Answer:

Earth's tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun's most direct rays. So, when the North Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it's winter in the Northern Hemisphere.

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3 years ago

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the

My name is Ann [436]

Answer:

2.82 s

Explanation:

The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 is the starting position, 2.3 m in this case.

Vy0 is the starting speed, 13 m/s.

a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.

Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2

It will reach the ground when Y(t) = 0

0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2

-4.9 * t^2 + 13 * t + 2.3 = 0

Solving this equation electronically gives two results:

t1 = 2.82 s

t2 = -0.17 s

We disregard the negative solution. The ball spends 2.82 seconds in the air.

7 0

3 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

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3 years ago

The speed of sound in air at room temperature is about s=343 m/s. calculate this speed in miles per hour (mph). (1 mi = 1609 m.)

igor_vitrenko [27]

Given:
1 mi = 1609 m.

Also,
1 hour = 3600 s

Therefore
$343 \, \frac{m}{s} =(343 \, \frac{m}{s} )*(3600 \, \frac{s}{h}) *( \frac{1}{1609}\, \frac{mi}{m})= 767.433 \, \frac{mi}{h}$

Answer: 767.4 mph (nearest tenth)

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4 years ago

Read 2 more answers