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vivado [14]
3 years ago
8

Chemical, gravitational, and elastic are examples of types of what? A. systems B. potential energy C. kinetic energy D. engines

Physics
1 answer:
murzikaleks [220]3 years ago
4 0

Chemical, Gravitational, and Elastic are examples of: B. Potential Energy. Hope it helps.

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A man weighing 180 lbf pushes a block weighing 100 lbf along a horizontal plane. the dynamic coefficient of friction between the
Talja [164]
The first thing you should know is that the work is defined as:
 W = F * d
 Where
 F = force
 d = displacement
 We have then
 (a) the block
 F = (0.2) * (100) = 20
 d = 100
 W = (20) * (100) = 2000 ft.lbf
 (b) the man as the system.
 F = (0.2) * (100 + 180) = 56
 d = 100
 W = (56) * (100) = 5600 ft.lbf
 answer:
 (a) 2000 ft.lbf
 (b) 5600 ft.lbf
3 0
3 years ago
What's the formula to find out power
Julli [10]

In general, 

                 Power = (energy moved) / (time to move the energy) .

If it's mechanical power, then     

                 Power = (work done) / (time to do the work) .

If it's electrical power, then it can be any one of these:

                 Power  =  (volts)  x  (amperes)

                 Power  =  (volts)²  /  (resistance, ohms)

                 Power  =  (amperes)²  x  (resistance, ohms) .

Whatever kind of energy you're dealing with, power always
turns out to be

                  (amount of energy produced, used, or moved)
divided by
                  (time taken to produce, use, or move the energy) .          
3 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
Everyday activities do NOT produce significant muscle growth because:
Reil [10]

Answer:

When elements bond together or when bonds of compounds are broken and form a new substance

Explanation:

7 0
3 years ago
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What are some ways to conserve electricity
emmasim [6.3K]
Turn off lights when leaving rooms.

Unplug unused appliances. Even when powered off these appliances use electricity.

Replace regular light bulbs with energy saving bulbs.
7 0
3 years ago
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