Answer:
a) 19440 km/h²
b) 10 sec
Explanation:
v₀ = initial velocity of the car = 45 km/h
v = final velocity achieved by the car = 99 km/h
d = distance traveled by the car while accelerating = 0.2 km
a = acceleration of the car
Using the kinematics equation
v² = v₀² + 2 a d
99² = 45² + 2 a (0.2)
a = 19440 km/h²
b)
t = time required to reach the final velocity
Using the kinematics equation
v = v₀ + a t
99 = 45 + (19440) t
t = 0.00278 h
t = 0.00278 x 3600 sec
t = 10 sec
Because it is if you know you know and it is also helping the sentwcnde and air and confiscation
Answer:
Pe is stored energy to do work Pe is stored in objects that are at rest
KE defined as the energy of motion
began to move
Not sure about the rest
hope this helps
Explanation:
Answer:
a) Em₀ = 42.96 104 J
, b) 
= -2.49 105 J
, c) vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
Em = K + U
a) Let's look for the initial mechanical energy
Em₀ = K + U
Em₀ = ½ m v2 + mg and
Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
Em₀ = 36 104 + 6.96 104
Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
= Em₂ -Em₀
Em₂ = K + U
Em₂ = ½ m v₂² + m g y₂
Em₂ = ½ 50 85 2 + 50 9.8 427
Em₂ = 180.625 + 2.09 105
Em₂ = 1,806 105 J
= Em₂ -Em₀
= 1,806 105 - 4,296 105
= -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
= ΔEm
= Emf - Emo
-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
½ 50.0 vf² = 0.561
vf = √ 0.561 25
vf = 3.75 m / s
-- Accelerating at the rate of 8 m/s², Andy's speed
after 30 seconds is
(8 m/s²) x (30.0 s) = 240 m/s .
-- His average speed during that time is
(1/2) (0 + 240 m/s) = 120 m/s .
-- In 30 sec at an average speed of 120 m/s,
Andy will travel a distance of
(120 m/s) x (30 sec) = 3,600 m
= 3.6 km .
"But how ? ! ?", you ask.
How in the world can Andy leave a stop light and then
cover 3.6 km = 2.24 miles in the next 30 seconds ?
The answer is: His acceleration of 8 m/s², or about 0.82 G
is what does it for him.
At that rate of acceleration ...
-- Andy achieves "Zero to 60 mph" in 3.35 seconds,
and then he keeps accelerating.
-- He hits 100 mph in 5.59 seconds after jumping the light ...
and then he keeps accelerating.
-- He hits 200 mph in 11.2 seconds after jumping the light ...
and then he keeps accelerating.
-- After accelerating at 8 m/s² for 30 seconds, Andy and his
car are moving at 537 miles per hour !
We really don't know whether he keeps accelerating,
but we kind of doubt it.
A couple of observations in conclusion:
-- We can't actually calculate his displacement with the information given.
Displacement is the distance and direction between the starting- and
ending-points, and we're not told whether Andy maintains a straight line
during this tense period, or is all over the road, adding great distance
but not a lot of displacement.
-- It's also likely that sometime during this performance, he is pulled
over to the side by an alert cop in a traffic-control helicopter, and
never actually succeeds in accomplishing the given description.
