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julia-pushkina [17]
3 years ago
14

. While a person lifts a book of mass 2 kg from the floor to a tabletop, 1.5 m above the floor, how much work does the the perso

n do on the book? How much work does gravity do on the boo
Physics
2 answers:
viktelen [127]3 years ago
5 0

Answer:

a) 29.4 J

b) - 29.4 J

Explanation:

Given:

Mass of the book, m = 2 kg

Height above the floor, h = 1.5 m

Now,

the work done by the person will be = Force applied on the book × displacement of the book

thus,

Work done by the person = mg × h

where, g is the acceleration due to gravity

thus, on substituting the values, we get

Work done by the person = 2 × 9.8 × 1.5 = 29.4 J

now,

for the force applied by the gravitational pull (downwards) the displacement is in opposite direction (upwards) to the force of the gravity.

Thus,

work done by the gravity will be negative

therefore, the work done by the gravity = - mg × h

or

work done by the gravity = - 29.4 J

photoshop1234 [79]3 years ago
4 0

Answer:

29.4 J  -29.4 J

Explanation:

The work done by the person on the book is given by =mgh

where m =mass

          g= acceleration due to gravity

          h=height

Here m=2 kg given g=9.8 kg m/sec^2 h=1.5 meter

So work done by the person =2×9.8×1.5=29.4 J

As the gravitational force work downward and person lift the book upward so work done by gravity = -mgh = -2×9.8×1.5=  -29.4 J  

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Explanation:

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The magnetic flux through a coil of wire containing two loops changes at a constant rate from -83 Wb to 82 Wb in 0.39 s .
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Answer:

423v

Explanation:

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PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.
AleksAgata [21]
22. a - (vf^2 - vi^2)/(2d) 
a = (0 - 23^2)/(170) 
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3 
33 = 3t 
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long. 

24. The sprinter starts from rest. The average acceleration is found from: 
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m 

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s 
acceleration, a = -7.00m/s^2
displacement, s - 92m 
Using v^2 = u^2 - 2as 
0^2 - u^2 + 2 (-7.00) (92) 
initial velocity, u = sqrt (1288) = 35.9 m/s 
This is the speed pf the car just bore braking. 

I hope this helps!! 

5 0
3 years ago
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