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2 years ago

HELP PLEASE...

Physics

1 answer:

skad [1K]2 years ago

7 0

Ya because they’re are both 50 liters

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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

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

5 0

3 years ago

Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the

tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s$

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m$

The displacement of the ball on the moon is 7.04 m

6 0

3 years ago

Which describes the properties reflection, absorption, and transmission when a light is shinning on the oval end of a silver spo

Trava [24]

The correct answer is C. Hope this helps

8 0

3 years ago

The atomic number of an element is found by?

azamat

Calculate the number of neutrons. Now you know that atomic number = number of protons, and mass number = number of protons + number of neutrons. To find the number of neutrons in an element, subtract the atomic number from the mass number.

6 0

2 years ago

Read 2 more answers