Ask question

Ask question

All categories

2 years ago

11

The Earth's electric field creates a potential that increases 100 V for every meter of altitude. If an object of charge 4.5 mC a

nd mass 68 g falls a distance of 1.0 m from rest under the influence of the Earth's electric and gravitational fields, what is its final kinetic energy

1 answer:

castortr0y [4]2 years ago

8 0

Answer:

The final kinetic energy is $K = 1.1 \ J$

Explanation:

From the question we are told that

The electric field is $E = 100 \ V/m$

The charge on the object is $q = 4.5 mC = 4.5 *10^{-3} \ C$

The mass of the object is $m_o = 68 \ g = 0.68 \ kg$

The distance moved by the object is $d = 1.0 \ m$

The workdone on the object by the fields is mathematically represented as

$W = [qE + mg]d$

Now this workdone is equivalent to the final kinetic energy so

$K = W = [qE + mg]d$

substituting values

$K = W = [4.5*10^{-3 } *100 + 0.68 * 9.8]* 1$

$K = 1.1 \ J$

You might be interested in

A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At

NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

$B=\frac{1}{2} \mu \epsilon_0 r$

Where,

$\mu =$ Permeability constant

$\epsilon_0 =$ Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

$B \propto r$

Then the distance relationship would be given by

$\frac{r}{R} = \frac{B}{B_{max}}$

$r =\frac{B}{B_{max}} R$

$r = \frac{0.5B_{max}}{B_{max}}R$

$r = 0.5R$

On the outside, however, it is defined by

$B = \frac{\mu_0 i_d}{2\pi r}$

Here the magnetic field is inversely proportional to the distance, that is

$B \not\propto r$

Then,

$\frac{r}{R} = \frac{B_{max}{B}}$

$r = \frac{B_{max}{B}}R$

$r = \frac{B_{max}{0.5B_{max}}}R$

$r = 2R$

7 0

3 years ago

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x\\$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

5 0

3 years ago

The graph above shows the position and time calculate the velocity of the particle from T=0s to T=4s?

zzz [600]

Answer:

The velocity of the particle from T = 0 s to T = 4 s is;

0.5 m/s

Explanation:

The given parameters from the graph are;

The initial displacement (covered) at time, t₁ = 0 s is x₁ = 1 m

The displacement covered at time, t₂ = 4 s is x₂ = 3 m

The graph of distance to time, from time t = 0 to time t = 4 is a straight line graph, with the velocity given by the rate of change of the displacement to the time which is dx/dt which is also the slope of the graph given as follows;

$The \ slope \ of \ the \ displacement \ time \ graph, \ m =velocity, \ v= \dfrac{x_2 - x_1}{t_2 - t_1}$

$Therefore, \ the \ velocity \ of \ the \ particle \ v = \dfrac{3 \ m - 1 \ m}{4 \ s - 0 \ s} = \dfrac{2 \ m}{4 \ s} = \dfrac{1}{2} \ m/s$

The velocity of the particle from t = 0 s to t = 4 s = 1/2 m/s = 0.5 m/s.

3 0

2 years ago

Heat energy refers to the kinetic energy of molecules. heat can move in a number of different ways: when warm air rises causing

Anna71 [15]

Convection is the process that is said to occur

7 0

3 years ago

Read 2 more answers

An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N, find the coefficient of kinetic

Stolb23 [73]

Answer:

The coefficient of kinetic friction = 0.026

Explanation:

An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N.

Here we have to note that the weight is pulled at a constant speed . This means that the net force acting on the weight is zero.

The external force acting on the body is in the forward direction and the friction acts in the backward direction.

Friction increases as the mass of the body increases.

Friction = $u_{k}\times m \times g$

We now equate this to the external force of 15 N.

15 = $u_{k} \times 56 \times 10$

$u_{k}$ = $\frac{15}{560}$

$u_{k}$ = 0.026

The coefficient of kinetic friction = 0.026

8 0

3 years ago