Which examples are simple machines?

You are on roller blades on top of a small hill. Your potential energy is equal to 1,000.0 joules. The last time

irinina [24]

It is -59 i think or just ask someone esle for help

8 0

3 years ago

What is a suspension bridge- in your own words please and ty

PIT_PIT [208]

A bridge supported by vertical cables which then leads to more support from larger cables.

4 0

3 years ago

if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?

madreJ [45]

Answer:

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

$a_{c} =\frac{(velocity)^{2} }{radius}$

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

$a_{c} = ?$

Solution:

We have,

$a_{c} =\frac{(velocity)^{2} }{radius}$

Substituting given value in it we get

$a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2$

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

7 0

3 years ago

A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal

Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

$a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2$

Let d be the breaking distance. It can be calculated using third equation of motion as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m$

So, the required distance covered by the car is 44.64 m.

4 0

3 years ago

A small rock is thrown vertically upward with a speed of 21.0 m/sm/s from the edge of the roof of a 21.0-mm-tall building. The r

FromTheMoon [43]

Answer:

Explanation:

A.

Given:

Vo = 21 m/s

Vf = 0 m/s

Using equation of Motion,

Vf^2 = Vo^2 - 2aS

S = (21^2)/2 × 9.8

= 22.5 m.

B.

Given:

S = 22.5 + 21 mm

= 22.521 m

Vo = 0 m/s

Using the equation of motion,

S = Vo × t + 1/2 × a × t^2

22.521 = 0 + 1/2 × 9.8 × t^2

t^2 = (2 × 22.521)/9.8

= 4.6

t = 2.14 s

4 0

4 years ago