How to calculate F2?<br> m=16.4kg<br> f1= 2.7n<br> angle=34.4

A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin

pychu [463]

a) See graph in attachment

b) The suvat equation to use is $v_f^2 - v_i^2 = 2as$

c) The acceleration is $a=\frac{v_f^2-v_i^2}{2s}$

d) The acceleration is $1.25 m/s^2$

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is $x_i = 0$ , the origin

- The final position of the boat is $x_f = 200 m$

- The initial velocity of the boat is $v_i = 20.0 m/s$

- The final velocity of the boat is $v_f = 30.0 m/s$

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

$v_f^2 - v_i^2 = 2as$

where

$a$ is the acceleration

c)

Now we have to solve the equation

$v_f^2 - v_i^2 = 2as$

In order to find the acceleration.

This can be done by dividing both terms by $2s$ : this way, we find

$\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}$

And so the acceleration is

$a=\frac{v_f^2-v_i^2}{2s}$

d)

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

And substituting into the equation,

$a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2$

e)

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

$v_f = v_i + at$

where:

$v_i$ is the initial velocity

$v_f$ is the final velocity

a is the acceleration

t is the time

Here we have:

$v_i = 20.0 m/s$

$v_f = 30.0 m/s$

$a=1.25 m/s^2$

Solving for t, we find:

$t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago

The football player running towards the goal line has

blagie [28]

the football player has speed

8 0

3 years ago

Read 2 more answers

Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of

IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0

3 years ago

When a burning stick of increase is moved fast in a circle a circle of red light is seen.

anzhelika [568]

Answer:

The impression of the image on the retina lasts for about 1/16th of a second after the removal of the object. If a burning stick of incense is revolved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Explanation:

7 0

2 years ago

a car accelerates uniformly from rest to a speed of 51.9mi/h in 9.37s. find the constant acceleration of the car. answer in unit

Darina [25.2K]

Here is my step-by-step-work. Let me know if you have any questions! :)

3 0

2 years ago

How to calculate F2? m=16.4kg f1= 2.7n angle=34.4