Ask question

Ask question

All categories

3 years ago

10

A car is moving 14.4 m/s when it hits the brakes. It slows for 78.8 m, which takes 9.92 s. It then accelerates at 1.83 m/s^2 for

4.50 s. What is its final velocity?
2-part motion
1-D kinematics

1 answer:

Cloud [144]3 years ago

6 0

This is a 2-part motion.
Part 1 of the motion:
Applying equations of motion;
u = 14.4 m/s, s = 78.8 m, t = 9.92 s

s = ut+1/2at^2
78.8 = 14.4*9.92 + 1/2*a*9.92^2
78.8 = 142.848 + 49.2032a
a = (78.8-142.848)/49.2032 = -1.3 m/s^2 (negative sign indicates a deceleration).
Now,
v = u+at
v = 14.4 -1.3*9.92 = 1.49 m/s

Part 2 of the motion:
Final speed in first motion becomes initial velocity in this second motion.
u = 1.49 m/s, a = 1.83 m/s^2, t = 4.50 s

Now,
v = u + at
Then,
v = 1.49 + 1.83*4.50 = 9.72 m/s

The final speed is 9.72 m/s

You might be interested in

Three long, straight wires are carrying currents that have the same magnitude. In C the current is opposite to that in A and B.

Nadusha1986 [10]

Answer:

(b) B

Explanation:

The direction of force on a current carrying wire in a magnetic field can be found using the right hand rule, which states that-"stretch the thumb in the direction of the current, and point the fingers in the direction of magnetic field. The direction of palm will then give the direction of force on the wire

On wire B the forces due to A and C act in the same direction and so strengthen each other. they get added up because the forces act in the same direction.

on wires A and C the forces (due to B and C and A and B

respectively) act in opposite directions and therefore tend to cancel out.

5 0

4 years ago

A 10 kg brick and a 1 kg book are dropped in a vacuum. The force of gravity on the 10 kg brick is what?

Nadya [2.5K]

<span>10 times as much. Since F=m*a, and a is constant, the only thing that affects force is the mass.

In response to the below answer, the acceleration due to gravity does not change. The force due to gravity definitely DOES change depending on the mass of the object. Since the force is what the problem asks for, the answer is 10</span>

7 0

3 years ago

Read 2 more answers

A smoke detector is an electronic fire-protection device that automatically senses the presence of smoke as a key indicator of f

Colt1911 [192]

Answer:

Photoelectric-type alarms aim a light source into a sensing chamber at an angle away from the sensor. Smoke enters the chamber, reflecting light onto the light sensor; triggering the alarm.

Explanation:

nfpa.org is the website with theanswer

7 0

3 years ago

Why is it advised not to hold the thermometer by its bulb while reading it?

Fiesta28 [93]

Because your body heat might change the temperature

7 0

2 years ago

Read 2 more answers

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>

This is what Einstein proposed:

Light behaves like a stream of particles called photons with an energy $E$ :

$E=h.f$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi$ is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago