Answer:
1.) Displacement = 32.08m
2.) Velocity = 55km/h
3.) Acceleration = 94.29 km/h^2
Explanation:
Given that the train takes 35 minutes to cover the distance between the two towns. The train's average speed is 55 kilometers per hour. That is,
Time t = 35 minute
Convert it to hours by dividing it by 60
35/60 = 7/12 hours
Speed = 55 km/h
The displacement of the car is:
Displacement = 55 × 7/12 = 32.08 m
The velocity of the car will be 55 km/h
The acceleration of the car will be:
Acceleration = velocity/time
Substitute velocity and time into the formula
Acceleration = 55 ÷ 7/12
Acceleration = 94.29 km/h^2
The correct choice is ❝ A ❞
❖ whєn wє tαlk αвσut α pєndulum, ít gσєѕ thrσugh ❝ tσ αnd frσ ❞ mσtíσn, αlѕσ knσwn αѕ hαrmσníc mσtíσn. αѕ вєíng ín thє cαѕє σf α pєndulum, thєrє'ѕ α cσntínuσuѕ chαngє ín kínєtíc αnd pσtєntíαl єnєrgч.
❖ whєn thє вσв íѕ αt mєαn pσѕítíσn, ít hαѕ ítѕ mαхímum vєlσcítч, αnd thє nєt fσrcє αnd αccєlєrαtíσn αctíng σn ít íѕ zєrσ, thєrєfσrє ít hαѕ mαхímum kínєtíc єnєrgч αt thαt pαrtículαr ínѕtαnt.
❖ whєrє αѕ, íf thє вσв íѕ αt σnє σf thє єхtrєmє pσѕítíσnѕ, ítѕ vєlσcítч вєcσmєѕ zєrσ, αnd thє αccєlєrαtíσn αnd fσrcє αctíng σn ít íѕ mαхímum, ѕímcє ít hαѕ nσ vєlσcítч, thєrєfσrє ít hαѕ nσ kínєtíc єnєrgч αt thαt ínѕtαnt, thє σnlч єnєrgч ít hαѕ αt thαt tímє íѕ pσtєntíαl єnєrgч.
❖ ѕíncє thє mσvєmєnt íѕ cσntínuσuѕ, cσntínuσuѕ chαngє ín єnєrgч σccurѕ, вut thє ѕum σf вσth thє єnєgíєѕ rєmαínѕ cσnѕtαnt αt єvєrч ínѕtαnt.
Answer:
–8.35 m/s²
Explanation:
We'll begin by converting 104 km/h to m/s. This can be obtained as follow:
3.6 Km/h = 1 m/s
Therefore,
104 km/h = 104 km/h × 1 m/s / 3.6 Km/h
104 km/h = 28.89 m/s
Thus, 104 km/h is equivalent to 28.89 m/s.
Finally, we shall determine the deceleration of the car. This can be obtained as follow:
Initial velocity (u) = 28.89 m/s
Final velocity (v) = 0 m/s
Distance (s) = 50 m
Deceleration (a) =?
v² = u² + 2as
0² = 28.89² + (2 × a × 50)
0 = 834.6321 + 100a
Collect like terms
0 – 834.6321 = 100a
–834.6321 = 100a
Divide both side by 100
a = –834.6321 / 100
a = –8.35 m/s²
Thus, the deceleration of the car is –8.35 m/s².
