A device that converts electrical energy into kinetic energy to turn an axle?

A boat travels at 15 m/s in a direction 45° east of north for an hour. The boat then turns and travels at 18 m/s in a direction

Nuetrik [128]

Answer:

brainly.com/question/5055678

Explanation:

4 0

3 years ago

Which factors affect the strength of the electric force between two objects

iris [78.8K]

By definition, the electric force is given by:

$f = k * \frac{q1q2}{d ^ 2}$

Where,

q1: electric charge of object number 1.

q2: electric charge of object number 2.

d: distance between both objects

k: proportionality constant

Therefore, the magnitude of the electric force is affected by:

1) The product of the charges of the objects

2) The distance between objects

Answer:

The factors that affect strength of the electric force between two objects are:

1) The product of the charges of the objects

2) The distance between objects

5 0

4 years ago

Read 2 more answers

A natural water molecule (H2O) in its vapor state has an electric dipole moment of magnitude, p = 6.2 x 10-30 C.m. (a) Find the

Vika [28.1K]

Answer:

a $D = 3.9 *10^{-12} \ m$

b $\tau_{max} = 1.24 *10^{-25} \ N\cdot m$

c $W = 2.48 *10^{-25} J$

Explanation:

From the question we are told that

The magnitude of electric dipole moment is $\sigma = 6.2 *10^{-30} \ C \cdot m$

The electric field is $E = 2*10^{4} \ N/C$

The distance between the positive and negative charge center is mathematically evaluated as

$D = \frac{\sigma }{10 e}$

Where e is the charge on one electron which has a constant value of $e = 1.60 *10^{-19} \ C$

Substituting values

$D = \frac{6.20 *10^{-30}}{10 * (1.60 *10^{-19})}$

$D = 3.9 *10^{-12} \ m$

The maximum torque is mathematically represented as

$\tau_{max} = \sigma * E * sin (\theta)$

Here $\theta = 90^o$

This because at maximum the molecule is perpendicular to the field

substituting values

$\tau_{max} = 6.2 *10^{-30} * 2*10^{4} sin ( 90)$

$\tau_{max} = 1.24 *10^{-25} \ N\cdot m$

The workdone is mathematically represented as

$W = V_{(180)} - V_{0}$

where $V_{(180)}$ is the potential energy at 180° which is mathematically evaluated as

$V_{(180) } = - \sigma * E cos (180)$

Where the negative signifies that it is acting against the field

substituting values

$V_{(180) } = - 6.20 *10^{-30} * 2.0 *10^{4} cos (180)$

$V_{(180) } = 1.24*10^{-25} J$

and

$V_{(0)}$ is the potential energy at 0° which is mathematically evaluated as

$V_{(0) } = - \sigma * E cos (0)$

substituting values

$V_{(0) } = - 6.20 *10^{-30} * 2.0 *10^{4} cos (0)$

$V_{(0) } =- 1.24*10^{-25} J$

So $W = 1.24 *10^{-25} - [-1.24 *10^{-25}]$

$W = 2.48 *10^{-25} J$

6 0

4 years ago

The nucleus of an atom can be modeled as several protons and neutrons closely packed together. Each particle has a mass of 1.67

Lady bird [3.3K]

Explanation:

The nucleus of an atom can be modeled as several protons and neutrons closely packed together.

Mass of the particle, $m=1.67\times 10^{-27}\ kg$

Radius of the particle, $R=10^{-15}\ m$

(a) The density of the nucleus of an atom is given by mass per unit area of the particle. Mathematically, it is given by :

$d=\dfrac{m}{V}$ , V is the volume of the particle

$d=\dfrac{m}{(4/3)\pi r^3}$

$d=\dfrac{1.67\times 10^{-27}}{(4/3)\pi (10^{-15})^3}$

$d=3.98\times 10^{17}\ kg/m^3$

So, the density of the nucleus of an atom is $3.98\times 10^{17}\ kg/m^3$ .

(b) Density of iron, $d'=7874\ kg/m^3$

Taking ratio of the density of nucleus of an atom and the density of iron as :

$\dfrac{d}{d'}=\dfrac{3.98\times 10^{17}}{7874}$

$\dfrac{d}{d'}=5.05\times 10^{13}$

$d=5.05\times 10^{13}\ d'$

So, the density of the nucleus of an atom is $5.05\times 10^{13}$ times greater than the density of iron. Hence, this is the required solution.

7 0

4 years ago

Someone answer these questions PLEASEE

lara31 [8.8K]

7. First write down all the known variables while separating the values for each direction:

x-direction:

vix = 20m/s

vfx = 20m/s

x = 39.2m

y-direction:

viy = 0m/s

ay = -9.8m/s^2

y = ?

Based on the knowns, the first step is to calculate the time of flight from the x-direction as it will be the same as value for the y-direction. Find the correct kinematic equation to do so:

x = (1/2)(vix+vfx)t

(39.2) = (1/2)(20+20)t

1.96s = t

Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the y-direction:

y = viy*t + (1/2)ay*t^2

y = (0)(1.96) + (1/2)(-9.8)(1.96)^2

y = -18.82m (Value is negative because gravity constant was negative. It is the height reference that from the top of the building down, which is why it is negative. The sign can be ignored for this question.)

8. First write down all the known variables while separating the values for each direction:

x-direction:

x = 12m

vfx = 0m/s

vix = ?

y-direction:

y1 = 1.2m

y2 = 0.6m

viy = 0m/s;

ay = -9.8m/s^2

First find time in the y-direction as it would be the same value for the x-direction.

(y2 - y1) = viy*t + (1/2)ay*t^2

(-0.6) = (0)t + (1/2)(-9.8)t^2

t = 0.35s

Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the x-direction:

x = (1/2)(vix+vfx)t

(12) = (1/2)(vix+(0))(0.35)

68.6m/s = vix

6 0

3 years ago