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3 years ago

A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it’s acceleration

2 answers:

7 0

So the initial velocity is 15 m/s, the final velocity is 0 since it's at a complete stop and time is 10 seconds. Therefore:

$Acceleration=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}} =\frac{0m/s-15m/s}{10s-0s} =-1.5\frac{m}{s^2}$

Therefore, the acceleration is -1.5 m/s^2. The reason it's negative is due to the fact that the vector is going against it's original movement since it's decelerating.

____ [38]3 years ago

3 0

Hello!

A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it's acceleration ?

We have the following data:

Vi (initial velocity) = 15 m/s (starts)

Vf (final velocity) = 0 m/s (stop)

t (time) = 10 s

a (acceleration) = ? (in m/s²)

We apply the data to the formula of the hourly function of the velocity, let us see:

$V_f = V_i + a*t$

$0 = 15 + a*10$

$- 15 = 10\:a$

$10\:a = - 15$

$a = \dfrac{-15}{10}$

$\boxed{\boxed{a = - 1.5\:m/s^2}}\Longrightarrow(the\:car\:slows\:down)\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:

The acceleration is -1.5 m/s² (decelerate)

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$\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}$

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Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68

creativ13 [48]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is $B= 4.2 *10^ {-6}T$ , the direction is into the page

Explanation:

From the question we are told that

The current is $i = 12.0 A$

The radius of arc bc is $r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m$

The radius of arc da is $r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m$

The length of segment cd and ab is = $l = 10cm = \frac{10}{100} = 0.10 m$

The objective of the solution is to obtain the magnetic field

Generally magnetic due to the current flowing in the arc is mathematically represented as

$B = \frac{\mu_o I}{4 \pi r}$

Here I is the current

$\mu_o$ is the permeability of free space with a value of $4\pi *10^{-7}T \cdot m/A$

r is the distance

Considering Arc da

$B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta$

Where $\theta$ is the angle the arc da makes with the center from the diagram its value is $\theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad$

Now substituting values into formula for magnetic field for da

$B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]$

$= \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]$

$B_{da}= 12.56*10^{-6} T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

$B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta$

Substituting value

$B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]$

$B_{bc}= 8.37*10^{-6}T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

Then the net magnetic field would be

$B = B_{da} - B{bc}$