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Nataliya [291]
3 years ago
15

A car accelerates if it:

Physics
1 answer:
Sladkaya [172]3 years ago
7 0

Answer:

A because when you accelerate you move even faster

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A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an
posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle \theta=30°

We need to calculate the angle in radian

\theta=30\times\dfrac{\pi}{180}

\theta=0.523\ rad

We need to calculate the path length

Using formula of path length

Path\ length =angle\times radius

Path\ length=0.523\times5.9

Path\ length =3.09\ m

(b). Angle = 30 rad

We need to calculate the path length

Path\ length=30\times5.9

Path\ length=177\ m

(c). Angle = 30 rev

We need to calculate the angle in rad

\theta=30\times2\pi

\theta=188.4\ rad

We need to calculate the path length

Path\ length=188.4\times5.9

Path\ length =1111.56\ m

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0
3 years ago
What type of atom is K2CO3
My name is Ann [436]
<span>Potassium carbonate (K2C03) is white salt and is often </span>found damp. It is soluble in water which makes a strong concoction. Hope this helps.
4 0
3 years ago
Read 2 more answers
3 m/s north is an example of a(n) ____.
alexandr1967 [171]
Rate of speed (3 m/s north is three miles per second north, so it's a rate of speed)
4 0
3 years ago
Read 2 more answers
engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi
Archy [21]

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

where ;

m_1=mass of object 1\\v_1i=initial  velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2

Given in the question that;

m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?

Apply the formula as;

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

<u>57 m/s = v₂f { nearest whole number}</u>

4 0
3 years ago
Read 2 more answers
A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900
Whitepunk [10]
It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.
5 0
3 years ago
Read 2 more answers
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