What is an example of kinetic energy

Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2

aleksandr82 [10.1K]

Answer:

The curl is $0 \hat x -z^2 \hat y -4xy \hat z$

Explanation:

Given the vector function

$\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z$

We can calculate the curl using the definition

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|$

Thus for the exercise we will have

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|$

So we will get

$\nabla \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z$

Working with the partial derivatives we get the curl

$\nabla \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z$

6 0

3 years ago

A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, def

Valentin [98]

Answer:

0.2448 point²

Explanation:

1 gry = 1/10 line

1 line = 1/12 inch

=> 1 gry in inches = 1/10 * 1/12 = 1/120 inch

=> 1 inch = 120 gry

1 point = 1/72 inch

=> 1 inch = 72 points

Therefore,

120 gry = 72 points

=> 1 gry = 3/5 point

Therefore,

1 gry² = (3/5)² point²

1 gry² = 9/25 point²

This means that 0.68 gry² will be:

0.68 gry² = 0.68 * 9/25 point²

=> 0.68 gry² = 0.2448 point²

6 0

2 years ago

4. There is a car with a large mass and a car with a small mass. If you push both cars with the

svlad2 [7]

I think the small mass Sorry if I’m incorrect

3 0

3 years ago

A proton is placed in an electric field of intensity 700 N/C. What are the magnitude and direction of the acceleration of this p

olya-2409 [2.1K]

Answer:

Acceleration of proton will be $a=0.67\times 10^{11}m/sec^2$

Explanation:

We have given a proton is placed in an electric field of intensity of 700 N/C

So electric field E = 700 N/C

Mass of proton $m=1.67\times 10^{-27}kg$

Charge on proton $e=1.6\times 10^{-19}C$

So electric force on the proton $F=qE=1.6\times 10^{-19}\times 700=1.120\times 10^{-16}N$

This force will be equal to force due to acceleration of the proton

According to newton's law force is given by F = ma

So $1.67\times 10^{-27}\times a=1.120\times 10^{-16}$

$a=0.67\times 10^{11}m/sec^2$

So acceleration of proton will be $a=0.67\times 10^{11}m/sec^2$

6 0

3 years ago

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

$F = ma$

Where F is the force experience by a particle

m is the mass of the particle

a is the acceleration of the particle

And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

$F = qvB$

Where q is the charge of the particle

v is the velocity of the charge

B is the magnetic field the charge is under it influence

Now equating this two formulas

$ma = qvB$

Making a the subject we have

$a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is $i$ hence the direction would be in the $i$ direction

So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

$a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

$a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

$a = 0.003 * 3i(2i+3j+4k)$

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

$i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

So

$a = 0.003* 6 =0.018\ m/s^2$

8 0

2 years ago