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3 years ago

6

A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

wer to the nearest whole number. J

2 answers:

morpeh [17]3 years ago

6 0

Answer:

1225 J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

$PE=mgh$

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

$PE=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J$

disa [49]3 years ago

4 0

Answer:

1225J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

1225J

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An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pi

Harman [31]

Answer:

The length is $l = 8.6 \ m$

Explanation:

From the question we are told that

The frequencies of the two successive harmonics are $f_1 = 220 \ Hz$ , $f_2 = 240 \ Hz$

The speed of sound in the air is $v_s = 343 \ m/s$

Generally the frequency of a given harmonic is mathematically represented as

$f_n = \frac{n v }{2l}$

Here n defines the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

$220 = \frac{n v}{2l}$

and

$240 = \frac{(n+1) v}{2l}$

So

$\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220$

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3 years ago

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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3 years ago

evablogger [386]

Answer:

b

Explanation:

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4 years ago

On Earth, a brick has a mass of 10 kg and a weight of 5 lbs. What predictions could we make about the mass and weight of the bri

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Answer:

Mass remains constant but weight reduces

Explanation:

Mass is the amount of matter in an object so whether on moon or any other planet, it does not change despite the changes in acceleration.

Weight is a product of mass and acceleration due to gravity, expressed as W=mg where m is the mass, W is weight and g is acceleration. From the above formula, it is evident that when you decrease g, then W also decreases while m is constant. Similarly, when m is constant and g is increased then W also increases.

Therefore, for this case, since g decreases, the weight decreases but mass remains constant.

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