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tatuchka [14]
3 years ago
11

A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2

seconds. The car's average acceleration during this time interval is
Physics
2 answers:
garik1379 [7]3 years ago
6 0
U1= 30m/s
u2= 23m/s
t=2s
a=(u2-u1)/t
a=-7/2=-3.5 m/s^2
mihalych1998 [28]3 years ago
3 0
Use the equation Vf=Vi+A(t)
Rearrange it so A is isolated: A= Vf-Vi/t
A=(23-30)/2
A=-3.5m/s^2
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What are dot diagrams in Physical Science?​
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Answer:

If you mean Lewis dot diagrams, aka electron-dot diagrams, then these are diagrams that show the bonding between atoms of a molecule, and the lone pairs of electrons that may exist in the molecule.

Explanation:

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Learning Goal:To understand and be able to use the rules for determining allowable orbital angular momentum states.Several numbe
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Answer:

Explanation:

1) for a given n value the l value can be from 0 to n-1

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5 0
3 years ago
A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
Andre45 [30]

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

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initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

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v₂ = √( (58.8)² - 2×9.8×16.59 )

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v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

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