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Genrish500 [490]
3 years ago
5

E. A racing car has a mass of 710kg. It starts from rest and travels 120m in

Physics
1 answer:
Amanda [17]3 years ago
7 0
6300
N

Explanation:
We will employ the following equation of kinematics
x
=
1
2
a
t
2

which describes an object traveling in one dimension with a constant acceleration and an initial velocity of zero.
We know the displacement and the time elapsed so we may solve for the acceleration:
40.0
m
=
1
2
a
(
3.0
s
)
2

Solving for
a
yields:
a
=
8.89
m
s
2

Now, knowing the acceleration of the car as well as the mass of the car we can apply Newton's second law, which states
F
net
=
m
a

All we need to do is plug in our values for
m
and
a
:
F
net
=
710
kg
⋅
8.89
m
s
2
=
6311.9
N

Considering the fact that our final answer should have only two significant digits, we will round to the nearest hundred:
F
net
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(a) The velocity of the first ball before the collision with the second ball is 11.18 m/s.

(b) The final velocity of the two balls after the collision is determined as 5.59 m/s.

<h3>Speed of the block when pushed by the spring</h3>

The speed of the block when pushed by the spring is calculated as follows;

K.E = Ux

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx²/m

v² = (25 x 0.5²)/0.05

v² = 125

v = 11.18 m/s

<h3>Final velocity of the two balls after the collision</h3>

The velocity of the two balls after the collision is calculated as follows;

Pi = Pf

where;

  • Pi is initial momentum
  • Pf is final momentum

m1u1 + m2u2 = v(m1 + m2)

0.05(11.18) + 0.05(0) = v(0.05 + 0.05)

0.559 = 0.1v

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2 years ago
You hang a light in front of your house using an
Kaylis [27]

The magnitudes of the forces that the ropes must exert on the knot connecting are :

  • F₁ = 118 N
  • F₂ = 89.21 N
  • F₃ = 57.28 N

<u>Given data :</u>

Mass ( M ) = 12 kg

∅₂ = 63°

∅₃ = 45°

<h3>Determine the magnitudes of the forces exerted by the ropes on the connecting knot</h3><h3 />

a) Force exerted by the first rope = weight of rope

∴  F₁ = mg

     = 12 * 9.81 ≈  118 kg

<u>b) Force exerted by the second rope </u>

applying equilibrium condition of force in the vertical direction

F₂ sin∅₂ + F₃ sin∅₃ - mg = 0  ---- ( 1 )

where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 )  applying equilibrium condition of force in the horizontal direction

Back to equation ( 1 )

F₂ =  [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]

   = [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]

   = 89.21 N

<u />

<u>C ) </u><u>Force </u><u>exerted by the</u><u> third rope </u>

Applying equation ( 2 )

F₃ = ( F₂ cos∅₂ / cos∅₃ )

    = ( 89.21 * cos 63 / cos 45 )

    = 57.28 N

Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :

F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N

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