Ask question

Ask question

All categories

2 years ago

9

A 2000kg car experiences a braking force of 10000N and skids to a stop in 6 seconds. The speed of the car just before the brakes

were applied was
A) 45 m/s
B) 15 m/s
C) 1.2 m/s
D) 30 m/s
E) none of these
So far I got a= 5m/s^2 and u= 30m/s but I think the answer is 45 m/s but I do not know where to go from there.

1 answer:

mylen [45]2 years ago

4 0

Answer:

D. 30m/s

Explanation:

According to the equation of motion

v=u +at

v = Final velocity

u = initial velocity

a = acceleration

t = time taken

Since we don't know the acceleration of the body, we will find its acceleration using newtons first law of motion F = ma

10000 =2000a

a = 5m/s²

initial velocity = 0m/s since the body starts from rest

v is what we are looking for

t = 6seconds

Substituting this datas in the formula

v = 0+ 5(6)

v = 30m/s (D)

You might be interested in

One strategy in a snowball fight is to throw

faltersainse [42]

Answers:

a) $\theta_{2}=23\°$

b) $t=1.199 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=11.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=67\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(11.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(67\°))$ (9)

$x=9.043 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=45.99\°$

$\theta_{2}=22.99\° \approx 23\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(11.1 m/s)sin(67\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.085 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(11.1 m/s)sin(23\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=0.885 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.085 s - 0.885 s$

Finally:

$t=1.199 s$

7 0

2 years ago

A positive point charge (q = +5.45 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.49 m. A p

ruslelena [56]

Answer:

$r_{B} = 3.34\ m$

Solution:

As per the question:

Point charge, q = $5.45\times 10^{- 8}\ C$

Test charge, $q_{o} = 4.96\times 10^{- 11}\ C$

Work done by the electric force, $W_{AB} = - 9.05\times 10^{- 9}\ J$

Now,

We know that the electric potential at a point is given by:

$V = \frac{kqq'}{r}$

where

r = separation distance between the charges.

Also,

The work done by the electric force i moving a test charge from point A to B in an electric field:

$W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}$

$- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}$

$- 0.3719 = (\frac{1}{r_{B}} - 0.67}$

$r_{B} = \frac{1}{0.299} = 3.34\ m$

5 0

3 years ago

In order to walk barefoot on hot coals without hurting your feet

siniylev [52]

Before a person walks through burning coal, the person will make sure their feet are very wet. When they start walking on the coal, this moisture will evaporate and form a protective gas layer underneath the person's feet. You can see examples of this if you happen to drip some water on a hot stove or any very hot surface. The water will very easily glide around on top of a newly formed layer of air underneath it -- like air hockey pucks on an air hockey table. Note that when someone walks through burning coal, typically this is also done very quickly to prevent a great deal of exposure to possible harm. By walking quickly, thinking positively, and letting the water cushion you from immediate danger over a short distance, such a task is possible. You may have also heard of physics teachers demonstrating how this principle works by sticking their hand first in a bucket of water and then quickly in a bucket of boiling molten lead. In the lead, their hand is protected briefly by a layer of gas from the evaporated water (the water vapor). I'm fairly sure that there is a name for this particular layer of gas, but I'm afraid the name is beyond me at the moment. In other words, water vapor has a low heat capacity and poor thermal conduction. Very often, the coals or wood embers that are used in fire walking also have a low heat capacity. Sweat produced on the bottom of people's feet also helps form a protective water vapor. All of this together makes it possible, if moving quickly enough, to walk across hot coals without getting burned. WARNING: Do not attempt to perform any of the actions described above. You can seriously injure yourself. Answered by: Ted Pavlic, Electrical Engineering Undergrad Student, Ohio St. (citing my source)

5 0

3 years ago

Assuming no air resistance, how far will a 0.0010 kg raindrop have fallen when it hits the ground 30.0 s later. Assume g = 9.8 m

Serga [27]

D = (1/2)·at²
where d is the distance fallen, a is the acceleration (g in this problem), and t is the time
d = (1/2)·(9.8 m/s²)·(30 s)² = (1/2)·(9.8)·(900) m
d = 4410 m
The answer is b) 4410 m

Note: the mass of the raindrop is irrelevant since the acceleration due to gravity is independent of mass. (Galileo's Leaning Tower of Pisa experiment)

6 0

3 years ago

Read 2 more answers

A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s,

Andreyy89

Answer:

-450 m/s

Explanation:

Momentum is conserved.

p₀ = p

0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v

v = -450 m/s

6 0

3 years ago