All categories

2 years ago

At what point on the y-axis is the resultant electric field of the two lines of charge equal to zero

Physics

1 answer:

Alja [10]2 years ago

3 0

answer it will help you to the slsfbsfgcxcjfgkiyyogikyugtgukyfjrasitfhtwalitfitkrursjyerjifykrtik ei

You might be interested in

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find

butalik [34]

Answer:

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy ..............................1

so it can be written as

work = force × distance ...................2

and

KE is express as

K.E = 0.5 × m × v²

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)² ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² = 919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

5 0

3 years ago

An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto

kupik [55]

Answer: rp/re= me/mp= 544 * 10^-6.

Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.

Then the dynamic equation for the circular movement is given by:

Fcentripetal= m*ω^2.r

q*v*B=m*ω^2.r

we write this for each particle then we have the following:

q*v*B=me* ω^2*re

q*v*B=mp* ω^2*rp

rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6

4 0

3 years ago

Velocity hdjsksskskskdldlddld

Nostrana [21]

Answer:

Vro....... really vro?

4 0

2 years ago

Ax = 44.4 m and Ay = 25.1 m Find the magnitude of the vector.

max2010maxim [7]

Answer:

The magnitude of the vector A is 51 m.

Explanation:

Given:

The horizontal component of a vector A is given as:

$A_x=44.4\ m$

The vertical component of a vector A is given as:

$A_y=25.1\ m$

Now, we know that, a vector A can be resolved into two mutually perpendicular components; one along the x axis and the other along the y axis. The magnitude of the vector A can be written as the square root of the sum of the squares of each component.

Therefore, the magnitude of vector A is given as:

$|\overrightarrow A|=\sqrt{A_{x}^2+A_{y}^2}$

Now, plug in 44.4 for $A_x$ , 25.1 for $A_y$ and solve for the magnitude of A. This gives,

$|\overrightarrow A|=\sqrt{(44.4)^2+(25.1)^2}\\|\overrightarrow A|=\sqrt{1971.36+630.01}\\|\overrightarrow A|=\sqrt{2601.37}\\|\overrightarrow A|=51\ m$

Therefore, the magnitude of the vector A is 51 m.

6 0

3 years ago

A 85.0kg man and a 65, 0kg woman are riding a Ferris wheel with a radius of 20.0m. What is the Ferris wheels tangential velocity

zvonat [6]

Answer:

The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is 3.92 m/s.

Explanation:

Let's use Newton's 2nd Law to help solve this problem.

F = ma

The force acting on the Ferris wheel is the centripetal force, given in the problem: $F_c=115 \ \text{N}$ .

The mass "m" is the sum of the man and woman's masses: $85+65= 150 \ \text{kg}$ .

The acceleration is the centripetal acceleration of the Ferris wheel: $a_c=\displaystyle \frac{v^2}{r}$ .

Let's write an equation and solve for "v", the tangential (linear) acceleration.

$\displaystyle 115=m(\frac{v^2}{r} )$
$\displaystyle 115 = (85+65)(\frac{v^2}{20})$
$\displaystyle 115=150(\frac{v^2}{20} )$
$.766667=\displaystyle(\frac{v^2}{20} )$
$15.\overline{3}=v^2$
$v=3.9158$

The Ferris wheel's tangential velocity is 3.92 m/s.

8 0

1 year ago