Ask question

Ask question

All categories

2 years ago

12

motorcycle accelerates from rest to a final velocity of 27 m/s in 9 seconds. What is its acceleration?

1 answer:

Vedmedyk [2.9K]2 years ago

5 0

Answer:

3 m/s squared

Explanation:

The formula you use is Vf= Vi + at. You rearrange it to a= Vf - Vi/t. The Vf is 27m/s. The Vi is 0m/s and the t is 9s. Cross out Vi since it’s zero and you’re left with a= 27m/s divided by 9s, which equals 3

You might be interested in

A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.

tankabanditka [31]

Answer:

$v_o=40.14\ m/s$

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

The horizontal speed is constant and equal to the initial speed $v_o$
The vertical speed is zero at launch time, but increases as the object starts to fall
The height of the object gradually decreases until it hits the ground
The horizontal distance where the object lands is called the range

We have the following formulas

$\displaystyle v_x=v_o$

$\displaystyle x=v_o.t$

$\displaystyle v_y=g.t$

$\displaystyle y=\frac{gt^2}{2}$

Where $v_o$ is the initial horizontal speed, $v_y$ is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

$\displaystyle y=\frac{gt^2}{2}$

Rearranging and solving for t

$\displaystyle 2y=gt^2$

$\displaystyle t^2=\frac{2\ y}{g}$

$\displaystyle t=\sqrt{\frac{2\ y}{g}}$

We then replace this value in

$\displaystyle x=v_o.t$

To get

$\displaystyle v_o=\frac{x}{t}$

$\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}$

$\displaystyle v_o=\sqrt{\frac{g}{2y}}.x$

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing $v_o$

$\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6$

The launch velocity is

$\boxed{v_o=40.14\ m/s}$

7 0

2 years ago

When an object loses electrons, it acquires a(n) ____________________ electric charge?

madam [21]

It acquires a positive electric charge.

4 0

3 years ago

Dans car has broken down. He decides to push it to a garage 800m away qlong a flat road. He pushes the car with qn average force

Ede4ka [16]

Answer:

Answer = 480000 joule.

mark me the Brainliest.. pls and follow me

3 0

2 years ago

An element's atomic number is 34. How many protons would an atom of this element have?

Elena-2011 [213]

34. The element would have 34 protons. Also 34 electrons.

7 0

2 years ago

A biker pedals hard to ride his bike to the top of a 44 m hill. He decides to let his bike coast down the hill, and is having so

Dafna11 [192]

Answer:

The bikers speed at the top of other hill is <u>25.82 m/s.</u>

Explanation:

Considering the biker is riding on a frictionless surface.

∴ There is no non-conservative or external force acting on the biker.

Hence we can conserve the energy of biker and bike as a system.

Let,

$h_{1}$ = 44m

$h_{2}$ = 10m

Since the biker starts from rest , his initial speed $v_{1}$ = 0 m/s

Let final speed of the bike at the top of other hill be $v_{2}$ .

∴ Initial Energy (at the top of 44m hill) = $mgh_{1}$

Final Energy (at the top of 10m hill) = $mgh_{2} + \frac{1}{2}mv_{2} ^{2}$ .

Conserving both the energies , we get

$mgh_{1}$ = $mgh_{2} + \frac{1}{2}mv_{2} ^{2}$

∴ $v_{2} = \sqrt{2g(h_{1}-h_{2} )}$

Substituting the values for g , $h_{1}$ , $h_{2}$ , we get

$v_{2}$ = 25.82 m/s

6 0

2 years ago