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Archy [21]
2 years ago
14

Pls answerrr it's urgent!!!!!

Physics
1 answer:
lilavasa [31]2 years ago
6 0

Explanation:

the average velocity of the car is 15 m/s example I have this on a test

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the radius of the entire atom was 0.00000001 cm.

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A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

acceleration; a = 2 m/s²

Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

6 0
3 years ago
In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)

<u>v₂ = 6.4 m/s</u>

The final speed of the leading car is given by the following formula:

v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)

<u>v₁ = 3.5 m/s</u>

4 0
3 years ago
NEED ASAP PLEASE !!
Svetllana [295]
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