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zaharov [31]
3 years ago
9

A man pulls on his dog's leash to keep him from running after a bicycle. Which term best describes this example? Select one: A.

negative work B. positive work C. negative force D. positive force
Physics
1 answer:
madreJ [45]3 years ago
6 0
C. Negative force. The dog isn't going to learn that way.

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navik [9.2K]

Answer:I think its the first one

Explanation:

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Electrical Safety: What should you do to a hot plate before turning it on? More than one answer may be correct. Check that the l
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Answer and Explanation:

Before turning a hot plate on, one should consider these points:

  • Check for the leads and ensure they does not trail in liquid.
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3 years ago
Which of the following statements are true? Positively charged objects attract other positively charged objects. Negatively char
butalik [34]
The statement that is true is that positively charged objects attract negatively charged objects. This is due to a law that states 'like forces attract while unlike forces repel. This same concept applies to magnetism. If you put two similar poles together, for example; if you place two south poles together. You feel a separating force between the two poles. But if you place two opposite poles together they attract each other. Hope i helped. <span />
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a person expended 500 newtons to move a full wheelbarrow 30 meters how much work was done? a person expended 500 newtons to move
stiv31 [10]
<span>A person expended 500 newtons to move a full wheelbarrow 30 meters how much work was done?

</span>In physics, a force is said to do work<span> if, when acting, there is a displacement of the point of application in the direction of the force. The work done would be calculated as follows:
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W = F x d = 500 N (30 m) = 15000 J or 15 kJ
6 0
3 years ago
A proton with mass 1.67*10^-27kg is propelled at an initialspeed of 3.00*10^5m/s directly toward a uranium nucleus 5.00maway. Th
Tresset [83]

Answer:

Explanation:

F = 2.12 x 10⁻²⁶ / x²

Work done by electric field of nucleus

W = ∫ Fdx

= ∫2.12 x 10⁻²⁶ / x² dx

= 2.12 x 10⁻²⁶ ( - 1 / x )

= - 2.12 x 10⁻²⁶ ( 1/5 - 1 / 8 x 10⁻¹⁰ )

= - .265 x 10⁻¹⁶ J

1/ 2 x mv² = .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ - .265 x 10⁻¹⁶

= 7.515 x 10⁻¹⁷ - .265 x 10⁻¹⁶

=( .7515 - .265 )x 10⁻¹⁶

= .4865 x 10⁻¹⁶

.5 x 1.67 x 10⁻²⁷ x v² = .4865 x 10⁻¹⁶

v² = .5826 x 10¹¹

v² = 5.826 x 10¹⁰

v = 2.41 x 10⁵ m /s

b )

Let r be the closest distance

Potential at this point

2.12 x 10⁻²⁶ (  1 / r )

Kinetic energy

= 0

Total energy = 2.12 x 10⁻²⁶ (  1 / r )

Total energy at 5 m

= .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ + 0 ( potential energy at 5 m will be negligible as compared with that near the center )

= 7.515 x 10⁻¹⁷ J

So ,

2.12 x 10⁻²⁶ (  1 / r ) = 7.515 x 10⁻¹⁷

r = 2.12 x 10⁻²⁶ / 7.515 x 10⁻¹⁷

= .282 x 10⁻⁹

= 2.82 x 10⁻¹⁰ m

c ) As electric field is conservative , no dissipation of energy takes place . Hence its speed at 5m on returning back to this point will be same as

3.00 x 10⁵ m /s

7 0
4 years ago
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