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3 years ago

Pls help guys this is my 7th time posting XD

Physics

2 answers:

borishaifa [10]3 years ago

5 0

Answer:

Explanation:

initial velocity u = 54 km/h = 15 m/s

final velocity v = 18 km/h = 5 m/s

distance s = 10 m

1. v^2 = u^2 + 2as

5^2 = 15^2 + 2a × 10

25 = 225 + 20a

25 - 225 = 20a

20 a = -200

a = -200/2

a = -100m/s^2

∴Deacceleration = -100m/s^2

2. v = u + at

5 = 15 -100t

5-15 = -100t

-10 = -100t

t = 100 / 10

∴t = 10 sec

Total distance covered by the car 10 m

zysi [14]3 years ago

5 0

To me, the hardest part of this whole thing is keeping the units straight. We're starting out with information given to us in kilometers, hours, and meters, and we have to come up with answers in m/s² , seconds, and meters.

When I worked this problem, I jumped right in without thinking, and I immediately got bogged down when I had to go off to the side and convert some units.

Now I know better. THIS time, before we get all tangled up trying to solve anything, let's get clever and change everything to m/s right now !

(54 km/hour) · (1,000m/km) · (1 hour/3600 sec) = 15 meters/second

(18 km/hour) · (1,000 m/km) · (1 hour/3600 sec) = 5 meters/second

NOW I'll betcha it's gonna be about 70% faster and easier !

i). Acceleration = (change in speed / time for the change)

We know the distance, but not the time. I know there's a formula for it, but I've learned so many formulas during my lifetime that I can't remember ALL of them. So I just memorize some of them, and I work things out from the formulas that I know. Here's how I do time:)

Average speed during the given slow-down = (1/2)·(15+5) = 10 m/s

Distance covered during the given slow-down = 10 m.

Time = (distance) / (average speed) = (10m) / (10 m/s) = 1 second

Acceleration = (change in speed) / (time for the change)

Acceleration = (5 m/s - 15 m/s) / (1 second)

Acceleration = (-10 m/s) / (1 second)

Acceleration = -10 m/s² (or 'Deceleration' = 10 m/s² )

_____________________________________________

For parts ii). and iii)., there's a big shift in the question.

It only gave you the slow-down from 54 to 18 km/hr for the purpose of calculating the deceleration. NOW, for the rest of the answers, it's talking about a complete stop ... 0 m/s .

____________________________________________

ii). Time required to stop = (initial speed) / (deceleration)

Time to stop from 54 km/hr = (15 m/s) / (10 m/s²)

Time to stop = 1.5 seconds

iii). Distance covered = (average speed) · (time to stop)

Distance covered = (7.5 m/s) · (1.5 sec)

Distance covered = 11.25 meters

OR ... use the official formula:

Distance = (1/2) · (acceleration) · (time² )

Distance = (1/2) · (10 m/s²) · (1.5 sec)²

Distance = 11.25 meters yay !

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The speed of a moving bullet can be deter-

Bad White [126]

Answer:

v = 381 m/s

Explanation:

Linear Speed

The linear speed of the bullet is calculated by the formula:

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

Where:

x = Distance traveled

t = Time needed to travel x

We are given the distance the bullet travels x=61 cm = 0.61 m. We need to determine the time the bullet took to make the holes between the two disks.

The formula for the angular speed of a rotating object is:

$\displaystyle \omega=\frac{\theta}{t}$

Where θ is the angular displacement and t is the time. Solving for t:

$\displaystyle t=\frac{\theta}{\omega}$

The angular displacement is θ=14°. Converting to radians:

$\theta=14*\pi/180=0.2443\ rad$

The angular speed is w=1436 rev/min. Converting to rad/s:

$\omega = 1436*2\pi/60=150.3776\ rad/s$

Thus the time is:

$\displaystyle t=\frac{0.2443\ rad}{150.3776\ rad/s}$

t = 0.0016 s

Thus the speed of the bullet is:

$\displaystyle v=\frac{0.61}{0.0016}$

v = 381 m/s

7 0

2 years ago

A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the

topjm [15]

Explanation:

This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

Let us denote each of the individual displacements by a vector. Consider the unit vectors $\vec{i}\textrm{ and }\vec{j}$ as the unit vectors in the direction of East and North respectively.

By simple calculations, we can derive the unit vectors $\vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2}$ in the directions North, $45^{o}$ South of West and $60^{o}$ North of West respectively.