1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Degger [83]
3 years ago
14

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?
Physics
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.

You might be interested in
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

3 0
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
DedPeter [7]

Answer:

Explanation:

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

7 0
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:

electric potential

I think so

6 0
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

8 0
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
Other questions:
  • A Savior In Physics Please Help In This Question!!!!!!!!!!!!!!!!
    15·1 answer
  • A satellite of mars, called phobos, has an orbital radius of 9.4 106 m and a period of 2.8 104 s. assuming the orbit is circular
    6·2 answers
  • Distance from any point on a wave to an identical point on the next wave?
    13·1 answer
  • What is the minimum total energy released when
    11·2 answers
  • PHYSICAL SCIENCE
    11·2 answers
  • Which of the following is the cause of the change of seasons? A) the distance of a place from the Equator B) prevailing winds bl
    5·1 answer
  • A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downwar
    7·1 answer
  • What happens to the current in a circuit if a 1.5 volt battery
    6·1 answer
  • A baseball pitcher throws a ball at 40.0 m/s towards home plate. The ball is thrown horizontally. How far will it drop on its wa
    15·1 answer
  • Which statement is true of a mechanical wave?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!