What is the sl unit for length

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

atroni [7]

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.

6 0

3 years ago

An electron in the beam of a TV picture tube is accelerated through a potential difference of 2.00 kv.?It then passes into a mag

Stells [14]

Answer:

The magnitude of the field is 8.384×10^-4 T.

Explanation:

Now, i start solving this question:

First, convert the potential difference(V) 2 kv to 2000 v.

As, we have the final formula is qvB = mv^2/r. It came from the centripetal force and the magnetic force and we know that these two forces are equal. When dealing with centripetal motion use the radius and not the diameter so

r = 0.36/2 = 0.18 m.

As, we are dealing with an electron so we know its mass is 9.11*10^-31 kg and its charge (q) is 1.6*10^-18 C.

We can solve for its electric potential energy by using ΔU = qV and we know potential energy initial is equal to kinetic energy final so ΔU = ΔKE and kinetic energy is equal to 1/2mv^2 J.

qV = 1/2mv^2

(1.6*10^-19C)(2000V) = (1/2)(9.11*10^-31kg) v^2

v = 2.65×10^7 m/s.

These all above steps we have done only for velocity(v) because in the final formula we have 'v' in it. So, now we substitute the all values in that formula and will find out the magnitude of the field:

qvB = mv^2/r

qB = mv/r

B = mv/qr

B = (9.11*10^-31 kg)(2.65×10^7 m/s) / (1.6*10^-19 C)(0.18 m)

Hence, B = 8.384*10^-4 T.

5 0

3 years ago

Which sentence uses the correct adverb to make a comparison?

Arturiano [62]

Option C would be the right one

8 0

3 years ago

A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so

shusha [124]

Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

$\displaystyle a=\frac{2\cdot 10}{5.12^2}$

$a=0.763\ m/s^2$

Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

$T_y+N=m.g$

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

$T_y=Tsin\theta$

$T_x=Tcos\theta$

Solving the above equation for N

$Tsin\theta+N=m.g$

$N=m.g-Tsin\theta\text{..........[1]}$

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

$Tcos\theta-F_r=m.a$