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Basile [38]
2 years ago
7

A car starts from rest and moves along the positive x-axis with a constant acceleration of 5 m/s/s for 8 seconds. It then contin

ues with constant velocity, i.e. the velocity it attains at 8 seconds for the next 5 seconds. What will be the distance covered by the car in the first 12 seconds of its motion?
Physics
2 answers:
boyakko [2]2 years ago
8 0
Answer is 340 plz believe
tino4ka555 [31]2 years ago
5 0

Answer:

340

Explanation:

Because you are moving at 5 m/s^2 for 8 seconds, you need to do some arithmetic to get the distance covered in the beginning. TO do this, you do 5 + 5*2 + 5*3 + 5*4 + 5*5 + 5*6 + 5*7 + 5*8, and once you have passed eight seconds, you need 4 more seconds with constant velocity, or 4(5*8). Your final equation would be 5+10+15+20+25+30+35+40+40+40+40+40. Hope this helps!

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You are designing a simple elevator system for an old warehouse that is being converted to loft apartments. A 22,500-N elevator
ICE Princess25 [194]

Answer:

A) mass = 3121.58 kg

B) tension = 25940.37 N

C) tension = 25940.37 N (tension on both sides will be the same)

Explanation:

Weight of elevator = 22500 N

Distance = 6.75 m

Time = 3 sec

Since it started from rest, initial speed is zero.

Using Newton's equation of motion we have,

S = ut + 0.5at^2

S = distance covered = 6.75 m

t = time = 3 s

a = acceleration upwards

u = initial velocity = 0

Substituting values, we have,

6.75 = 0(3) + (0.5 x a x 3^2)

6.75 = 4.5a

a = 6.75/4.5 = 1.5 m/s^2 (acceleration of the elevator upwards)

Mass of the elevator = Weight/g

Where g = acceleration due to gravity 9.81 m/s

Mass = 22500/9.81 = 2293.58 kg

From the image below we solve from

T - 22500 = ma

T - 22500 = 2293.58 x 1.5

T - 22500 = 3440.37

T = 3440.37 + 22500 = 25940.37 N (this is the tension on the rope)

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9.81m - 25940.37 = 1.5m

(9.81 - 1.5)m = 25940.37

8.31m = 25940.37

m = 3121.58 kg (mass of counter weight)

See image below

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