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3 years ago

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Imagine that a car is traveling from your left to your right at a constant velocity. Which two actions could the driver take tha

t may be represented as (a) a velocity vector and an acceleration vector both pointing to the right and then (b) changing so the velocity vector points to the right and the acceleration vector points to the left?

1 answer:

Alona [7]3 years ago

5 0

Answer:

(a). The speed of car will increases.

(b). The speed of car will decreases.

Explanation:

Given that,

A car is traveling from your left to your right at a constant velocity.

We need to find two actions could the driver take

Using given data

(a). If the velocity and acceleration of car to the right

It means the velocity and acceleration of car is in same direction then the speed of car will increases.

(b). If the velocity to the right and the acceleration to the left

It means the velocity and acceleration of car is in opposite direction then the speed of car will decreases.

Hence, (a). The speed of car will increases.

(b). The speed of car will decreases.

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While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

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What is the approximate size of Earth’s magnetic field? Where are Earth’s magnetic poles? Where is the magnet that causes Earth’

nataly862011 [7]

Earths poles are on opposite sides of the earth

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3 years ago

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The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7

stich3 [128]

Answer:

Electric field acting on the electron is 127500 N/C.

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Initial speed of electron, u = 0

Final speed of electron, $v=3\times 10^7\ m/s$

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}$

$a=2.25\times 10^{16}\ m/s^2$

According to Newton's law, force acting on the electron is given by :

F = ma

$F=9.1\times 10^{-31}\times 2.25\times 10^{16}$

$F=2.04\times 10^{-14}\ N$

Electric force is given by :

F = q E, E = electric field

$E=\dfrac{F}{q}$

$E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}$

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

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an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu

Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

$E1= k*q1/(h1-h2)..............(1)$

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1

$[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2$ \\

$E1=[tex]E= 359502+359502\\E=719004 V/m$ V/m[/tex]

similarly electric feild due to negative charge q2=-40

$[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2$ \\

$E2=359502V/m$

Total electric feild E at the aircraft is given as

$E= E1+ E2\\$ ...............(2)

Put values of E1 and E2 in equation2

$\\E=359502+359502\\E= 719004V/m\\$

therefore s Total electric feild E at the aircraft is E= 719004V/m

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Montano1993 [528]

Answer:

D

Explanation:

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