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nlexa [21]
1 year ago
14

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
Physics
2 answers:
ValentinkaMS [17]1 year ago
8 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

Maksim231197 [3]1 year 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

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MaRussiya [10]

Answer:

dz=19217687.07\ m

Explanation:

Given:

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<u>The relation between the change in pressure with height is given as:</u>

\frac{dP_{max}}{dz} =\rho_a.g_n

where:

dz = height in the atmosphere

g_n= standard value of gravity

<em>Now putting the respective values:</em>

\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8

dz=19217.687\ km

dz=19217687.07\ m

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0
1 year ago
Gas a bG1 5.22 0.0289G2 1.05 0.0388G3 2.31 0.0467G4 4.05 0.0310Based on the given van der Waals constants for four hypothetical
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1 year ago
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid
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Answer:

a

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b

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Explanation:

From the question we are told that

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      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        z - (5-z) =  0

=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

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and

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=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

7 0
2 years ago
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Answer:

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