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3 years ago

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You want to create an electric field vector E = < 0, 5 104, 0> N/C at location < 0, 0, 0>. Where would you place a p

roton to produce this field at the origin?

1 answer:

atroni [7]3 years ago

6 0

Answer:

1.696 × 10^(-7) m on the y axis.

Explanation:

We are given the electric field as;

E = < 0, 5 × 10⁴, 0> N/C

This is written in (x, y, z) co-ordinates. So it means that it lies on the y-axis.

So,

E = 5 × 10⁴ N/C in the y direction.

Formula for Electric field is;

E = kq/r²

where;

k is a constant with a value of 8.99 x 10^(9) N.m²/C²

q is charge on the proton = 1.6 × 10^(-19) C

r is the distance

Thus, making r the subject gives;

r = √(kq/E)

Plugging in the relevant values gives;

r = √(8.99 × 10^(9) × 1.6 × 10^(-19)/(5 × 10⁴))

r = 1.696 × 10^(-7) m on the y axis.

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Two resistors R1 = 3 Ω and R2 = 6 Ω are connected in parallel. What is the net resistance in the circuit?

gtnhenbr [62]

Answer:

"2Ω" is the net resistance in the circuit.

Explanation:

The given resistors are:

R1 = 3Ω

R2 = 6Ω

The net resistance will be:

⇒ $\frac{1}{R_{net}} =\frac{1}{R_1} +\frac{1}{R_2}$

On substituting the values, we get

⇒ $\frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}$

On taking L.C.M, we get

⇒ $\frac{1}{R_{net}} =\frac{2+1}{6}$

⇒ $\frac{1}{R_{net}} =\frac{3}{6}$

⇒ $\frac{1}{R_{net}} =\frac{1}{2}$

On applying cross-multiplication, we get

⇒ $R_{net}=2 \Omega$

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3 years ago

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

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Answer:

Explanation:

3.4 m/s due North, -1.1 m/s due East

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3 years ago

Work done in taking charge from one point of a conductor to is another point is called

Yuliya22 [10]

Answer:

⁸

Explanation:

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3 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

So

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

B)

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

C)

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

D)

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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4 years ago

Function do lappet which faced vultures serve in a temperate grassland

vfiekz [6]

Um this doesn't make since to me since you did not clearly state your awnser

7 0

3 years ago

Read 2 more answers