I need to know the answer to the question

A particle, whose acceleration is constant, is moving in the negative x direction at a speed of 4.91 m/s, and 12.9 s later the p

Zinaida [17]

Answer:

The particle’s velocity is -16.9 m/s.

Explanation:

Given that,

Initial velocity of particle in negative x direction= 4.91 m/s

Time = 12.9 s

Final velocity of particle in positive x direction= 7.12 m/s

Before 12.4 sec,

Velocity of particle in negative x direction= 5.32 m/s

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$a=\dfrac{v-u}{t}$

Where, v = final velocity

u = initial velocity

t = time

Put the value into the equation

$a=\dfrac{7.12-(-4.91)}{12.9}$

$a=0.933\ m/s^2$

We need to calculate the initial speed of the particle

Using equation of motion again

$v=u+at$

$u=v-at$

Put the value into the formula

$u=-5.321-0.933\times12.4$

$u=-16.9\ m/s$

Hence, The particle’s velocity is -16.9 m/s.

4 0

3 years ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

pishuonlain [190]

Answer:

$E_{total}=4.82*10^6N/C$

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

$E_{total}=E_1+E_2$

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

$E=\sigma/(2\epsilon_o)$

Then:

$E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C$

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0

3 years ago

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Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, $d=24.44\times 2=48.88\ m$

(a) Acceleration, $a=-4\ m/s^2$

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

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

$s=\dfrac{-(24.44)^2}{2\times -4}$

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, $a=-8\ m/s^2$

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

$s=\dfrac{-(24.44)^2}{2\times -8}$

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0

3 years ago

Determine the displacement of a plane that is uniformly accelerated from 66 m/s to 88 m/s in 12 sec

Llana [10]

Determined the displacement of a plane is 66 m:s

5 0

3 years ago

2. Which of the following is NOT true of work?

salantis [7]

Answer:

D

Explanation:

Work is not a vector but it is a scalar

3 0

2 years ago

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