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antiseptic1488 [7]
3 years ago
5

While the change in blank will remain the same during a collision, the force needed to bring an object to a stop can be blank if

the time if collision is blank
Physics
2 answers:
Ahat [919]3 years ago
6 0

Explanation :

There are two types of collision i.e. elastic and elastic collision.

  • Elastic collision : In this type of collision, the total momentum and the kinetic energy of the particles remains constant.
  • Inelastic collision : In this type of collision, only the momentum remains constant while there is some loss of kinetic energy occurs.

From Newton's second law,

F = m a

a is the rate of change of velocity.

F=m\dfrac{v}{t}

There is a inverse relation between the force and the time of collision.

The change in <em><u>momentum</u></em> will remain the same during a collision, the force needed to bring an object to a stop can be <em><u>increased</u></em> if the time of the collision is <u><em>decreased</em></u>.

deff fn [24]3 years ago
5 0

Answer:momentum,decreased and increased

Explanation:

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nirvana33 [79]

Answer:

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Explanation:

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As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

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Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) =  7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

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E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

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nikitadnepr [17]

Answer:

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