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nlexa [21]
3 years ago
13

Resuelve:

Physics
1 answer:
Arlecino [84]3 years ago
3 0

The work done is 1764 J

Explanation:

The work done in lifting an object is equal to the gain in gravitational potential energy of the object:

W=mg \Delta h

where

m is the mass of the object

g is the acceleration of gravity

\Delta h is the change in height of the object

For the object in this problem, we have

m = 100 kg

g=9.8 m/s^2

\Delta h = 3 m -120 cm = 3 m- 1.20 m = 1.80 m

Therefore, the work done is

W=(100)(9.8)(1.80)=1764 J

Learn more about work and potential energy:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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Answer:

by obtaining the total mass of the dimes present:

d = 27.22 g / dozen       the density of dimes

M = n * d = 5 dozen * 27.22 g / dozen = 126.1 g

4 0
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A 2.0-Coulomb bead is given 20 Joules of electric potential energy by lifting it from the "ground" to point A. If a second bead
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Answer:

40 J

Explanation:

c_1 = 4 C

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The potential energy is directly proportional to the charge of the particle

U\propto c

\dfrac{U_2}{U_1}=\dfrac{c_2}{c_1}\\\Rightarrow U_2=\dfrac{U_1c_2}{c_1}\\\Rightarrow U_2=\dfrac{20\times 4}{2}\\\Rightarrow U_2=40\ J

The potential energy expected is 40 J

4 0
3 years ago
Which is the largest and most dense of the terrestrial planets? A.Mercury B.Venus C.Earth D.Mars
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C.Earth is the largest and most dense of the terrestrial planets

7 0
3 years ago
1-D Kinematics, Constant Acceleration After falling a distance of 45.0 m from the top of a building, a box is landing on the top
Mrrafil [7]

Answer:

Explanation:

Given

Object fall from a height of s=45\ m

Considering initial velocity to be zero i.e. u=0

using

v^2-u^2=2as  

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2-0=2(9.8)\cdot 45

v=29.69\ m/s\approx 29.7\ m/s

(b)Average acceleration

After falling 45 m, object strike the car and comes to rest after covering a distance of 0.5 m

again using

v'^2-u'^2=2as'

here final velocity will be zero i.e.v'=0

initial velocity u'=v

0-(29.7)^2=2\cdot a\cdot 0.5

a=-882.09\ m/s^2

(c)time taken by it to stop

v'=u'+a't

0=29.7-882.09\cdot t

t=0.034\ s

5 0
3 years ago
A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that i
Greeley [361]

Answer:

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x  component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here,  is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

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The rebound speed V' = Vby - Vbx

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V' = 989.95 m/s

5 0
4 years ago
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