All categories

3 years ago

While eating lunch high up in a skyscraper, two construction workers calculate their gravitational potential

Physics

1 answer:

Maurinko [17]3 years ago

8 0

Answer:

The mass of the other worker is 45 kg

Explanation:

The given parameters are;

The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker

The mass of the lighter construction worker, m₁ = 90 kg

The height level of the lighter construction worker's location = h₁

The height level of the other construction worker's location = h₂ = 2·h₁

The gravitational potential energy, P.E., is given as follows;

P.E. = m·g·h

Where;

m = The mass of the object at height

g = The acceleration due to gravity

h = The height at which is located

Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker

We have;

P.E.₁ = P.E.₂

Therefore;

m₁·g·h₁ = m₂·g·h₂

h₂ = 2·h₁

We have;

m₁·g·h₁ = m₂·g·2·h₁

m₁ = 2·m₂

90 kg = 2 × m₂

m₂ = (90 kg)/2 = 45 kg

The mass of the other construction worker is 45 kg.

You might be interested in

What maximum force do you need to exert on a relaxed spring with a 1.2×104-n/m spring constant to stretch it 6.0 cm from its equ

Elena L [17]

Answer:

Maximum force will be equal to 720 N

Explanation:

We have given that spring constant $k=1.2\times 10^4N/m$

Maximum stretch of the spring x = 6 cm = 0.06 m

We have to find the maximum force on the spring

We know that spring force is given by

$F=kx=1.2\times 10^4\times 0.06=720N$

So the maximum force which is necessary to relaxed the spring will be eqaul to 720 N

6 0

4 years ago

A 20.0-kg traffic light hangs midway on a cable between two poles 30.0 meters apart. If the sag in the cable is 0.40 meters, wha

hammer [34]

Answer:

the tension in each side of the cable is 3677.57 N

Explanation:

given data

traffic light = 20 kg

cable between two poles = 30 m

sag in the cable = 0.40 m

solution

by the free body diagram

tan θ = $\frac{0.4}{15}$ .............1

θ = 1.527 °

and

tension = mg

The net force is along x - axis is express as

T2 cosθ = T1 cosθ .................2

so T2 - T1 ..............3

and

when we take it along y axis that is express as

( T1 + T2) sinθ = mg ...................4

so by equation 3 we put here

2 × T1 sin(1.527) = 20 × 9.8

T1 = 3677.57 N

7 0

4 years ago

What is the difference between accurate data and reproducible data?

vladimir1956 [14]

Accuracy is a general concept while precision is more of a mathematical concept.

4 0

3 years ago

Read 2 more answers

Find the values of the root mean square translational speed v, molecules in gaseous diatomic oxygen (O2), gaseous carbon dioxide

frez [133]

Answer:

539.19 m/s , 459.83 m/s , 2156.78 m/s

Explanation:

The formula for the rms speed is given by

$v_{rms}=\sqrt{\frac{3RT}{M}}$

Where, R be the gas constant, T be the absolute temperature and M be the molecular mass of the gas

R = 8.314 in SI units

T = 100°C = 100 + 273 = 373 K

For Oxygen:

M = 32 g = 0.032 kg

$v_{rms}=\sqrt{\frac{3\times 8.314\times 373}{0.032}}$

vrms = 539.19 m/s

For carbon di oxide:

M = 44 g = 0.044 kg

$v_{rms}=\sqrt{\frac{3\times 8.314\times 373}{0.044}}$

vrms = 459.83 m/s

For hydrogen:

M = 2 g = 0.002 kg

$v_{rms}=\sqrt{\frac{3\times 8.314\times 373}{0.002}}$

vrms = 2156.78 m/s

3 0

3 years ago

A runner starts from rest and in 2 s reaches a speed of 7 m/s. If we assume that the speed changed at a constant rate (constant

PtichkaEL [24]

Answer:

$v=3.5\frac{m}{s}$

Explanation:

The average speed is defined as:

$v=\frac{\Delta x}{\Delta t}$

Using the equations for uniformly accelerated motion, we calculate the runner's acceleration:

$a=\frac{v_f-v_o}{2}\\a=\frac{7\frac{m}{s}-0\frac{m}{s}}{2s}\\a=3.5\frac{m}{s^2}$

Now, we can calculate the distance that the runner travels:

$\Delta x=v_0t+\frac{at^2}{2}\\\Delta x=(0\frac{m}{s})(2s)+\frac{3.5\frac{m}{s}(2s)^2}{2}\\\Delta x=7m$

Finally, we calculate the runner's average speed:

$v=\frac{7m}{2s}\\v=3.5\frac{m}{s}$

4 0

3 years ago