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AleksAgata [21]
3 years ago
7

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

Physics
1 answer:
Mademuasel [1]3 years ago
4 0
Hope this answers your question!! Ask any help at anytime
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When driving a car on the road
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What is the mass of a truck in grams of it produces a force of 1500N while accelerating at a rate of 6 m/s²?​
aleksley [76]

Answer:

250,000

Explanation:

<h2> </h2>

<h2>formula = ( F=ma </h2>

  • F=1500N
  • a=6m/s^2
  • F= ma
  • m=?
  • 1500/6 = m
  • m=250 kg
  • 1kg =1000gm so 250kg =250,000gm
  • m =250×10^3 gm
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3 years ago
compare your previous result to the present hrf result are there any changes if yes explain your answer​
garri49 [273]

Answer:

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Explanation:

kai6417

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2 years ago
A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener
trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

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

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

U=(m)(9.8)(15)=147 m (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

U=K=\frac{1}{2}mv^2

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

147 m = \frac{1}{2}mv^2

And re-arranging for v, we find the velocity:

v=\sqrt{2\cdot 147}=17.1 m/s

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3 years ago
define the term change and state one negative change you may encounter as a student or as an employee in the future​
Brums [2.3K]

Answer:

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ID: 724 645 3790

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