An electric current in a metal consists of moving

A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i

Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

$v^2=u^2+2as$

$0^2=14.5^2+2\times a\times 18.25$

$a=-5.76m/sec^2$

We know that acceleration is given by

$a=\mu g$

$5.76=\mu\times 9.81$

$\mu =0.587$

So coefficient of friction will be 0.587

6 0

3 years ago

Of the three sciences, physics, chemistry, and biology, the most complex is

Agata [3.3K]

Answer:

I would probably say C to be completely honest

Explanation:

If you agree make sure to give me a like

6 0

2 years ago

ASAP please 25 points if right

garik1379 [7]

Answer:

1. The resistance of any physical object to any velocity

2. It continues in it's existing state of rest or uniform motion

3. Mass is a quantity that is solely dependent upon the inertia of an object.

5 0

3 years ago

A hydraulic lift raises 1140 kg car through a height of 2.4 m. What is the potential energy

Sladkaya [172]

1140x9.8x2.4= 26,812.8 significant figures Make it 27,000

5 0

2 years ago

An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e

frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

$Tsin\theta= \frac{mv^2}{r}$

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

$Tcos\theta = mg$

Matching both expressions with respect to the tension we will have to

$T = \frac{\frac{mv^2}{r}}{sin\theta}$

$T = \frac{mg}{cos\theta}$

Then we have that,

$\frac{\frac{mv^2}{r}}{sin\theta} = \frac{mg}{cos\theta}$

$\frac{mv^2}{r} = mg tan\theta$

Rearranging to find the velocity we have that

$v = \sqrt{grTan\theta}$

The value of the angle is 14.5°, the acceleration (g) is 9.8m/s^2 and the radius is

$r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}$

$r = \frac{11.1}{2}+2.41$

$r = 7.96m$

Replacing we have that

$v = \sqrt{(9.8)(7.96)tan(14.5\°)}$

$v = 4.492m/s$

Therefore the speed of each seat is 4.492m/s

6 0

3 years ago