Okay guys I need serious help here plz!!!

A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.10 s. The particle has a charge

mamaluj [8]

Answer:

The magnitude of the electric field is $8.6\times10^{2}\ N/C$

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

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

Put the value into the formula

$a=\dfrac{160-0}{2.10}$

$a=76.19\ m/s^2$

We need to calculate the magnitude of the electric field

Using formula of electric field

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

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

$E=\dfrac{a+g}{\dfrac{q}{m}}$

Put the value into the formula

$E=\dfrac{76.19+9.8}{0.100}$

$E=8.6\times10^{2}\ N/C$

The direction is upward.

Hence, The magnitude of the electric field is $8.6\times10^{2}\ N/C$

4 0

3 years ago

Read 2 more answers

What is 541.2 mg in grams

PIT_PIT [208]

Answer:

0.5142 grams.

Explanation:

I think

5 0

3 years ago

Read 2 more answers

create a poem that incorporates those ten words. Feel free to make it as silly as you like! MINIMUM of 6 lines with a MINIMUM of

Luba_88 [7]

I could make a poem for you if you actually gave the words...... what 10 words do i need to incorporate???☹︎

8 0

3 years ago

Using the strap at an angle of 31.0° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15.0

lora16 [44]

Answer:

Explanation:

The question is incomplete. Here is the complete question.

Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.

Check the diagram related to the question in the attachment below for better understanding.

The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.

The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).

Ty = 15sin31°

Ty = 7.73N

W = mass * acceleration due to gravity

W = 15.0*9.8

W = 147N

The normal force is therefore expressed as;

N = Ty + W

N = 7.73 + 147

N = 154.73N

3 0

3 years ago

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0

3 years ago