Answers
Answer:
Explanation:
The total work done by the wave is expressed as;
Workdone = Potential energy + Kinetic energy
Workdone = mgh + 1/2mv²
m is the mass = 77kg
g is the acceleration due to gravity = 9.8m/s²
v is the velocity = 8.2m/s
h is the height = 1.65m
Substitute into the formula;
Workdone = 77(9.8)(1.65) + 1/2(77)8.2²
Workdone = 1245.09 + 2588.74
Workdone = 3833.83Joules
Hence the amount of non conservative work done on the sofa is 3833.83Joules
Given:
Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kgWe know,
→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]
or,
[tex]= mgh +\frac{1}{2} mv^2[/tex]
By putting the values,
[tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]
[tex]= 1245.09+2588.74[/tex]
[tex]= 3833.83 \ Joules[/tex]
Thus the above approach is right.
Learn more about work done here:
https://brainly.com/question/24230840
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Please help me with this question guys.
Answers
Answer:
The average speed is 22.2 km/h
Explanation:
Average Speed
Given an object travels a total distance d and took a total time t, then the average speed is:
[tex]\displaystyle \bar v=\frac{d}{t}[/tex]
The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:
[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]
Then he drives d2=7 km at v2=43 km/h taking a time of:
[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]
The total time is
t=0.467 h + 0.163 h = 0.63 h
The total distance is
d = 7 km + 7 km = 14 km
The average speed is:
[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]
The average speed is 22.2 km/h
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What type of evidence does the writer of this passage use to support her claim?
statistics
statistics
an expert's opinion
an expert's opinion
examples
examples
the writer's opinion
Answers
Answer:
b
Explanation:
An arrow in a bow has 357 J of elastic potential energy How much Winette enere
Will the arrow have after it has been shot assuming there is no sir restoran
Answers
Answer:
357 J
Explanation:
The elastic potential energy of arrow in the stretched bow is 357 J.
The kinetic energy of the arrow after it has been shot is given by half of the product of the arrow's mass and velocity of the arrow.
Here there are no other forms of energy at play here. Only potential and kinetic energy.
As we know that in any system the energy is conserved accordingly the elastic potential energy of the arrow will be equal to the kinetic energy of the bow after it is released i.e., 357 J.