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3 years ago

What's an anime I should watch next?? WEEPS ONLY ¨ˆ¨

1 answer:

3 0

Well uhhh if you’re a sub watcher I totally recommend Classroom of the Elite- it’s in crunchy roll, or Soul eater- that’s what I’m currently watching. Or Aot Junior- it’s on Hulu.

You might be interested in

Which of these systems is an oscillator? A. A barrel rolling down the hill B. A child sitting on a swing a skater falling on the

Paladinen [302]

Answer:

Option B:

A child sitting on a swing.

Explanation:

When we hear the word oscillator, a good example is the pendulum bob of a grandfather clock. We can picture the motion to get a perfect understanding of its path of motion and relate it to other systems of motion in our everyday life.

An oscillator is a system that moves in such a way that it reverses its direction after a period of time. It can be seen as a "to-and-fro" motion.

From the options, a child sitting on a swing is the perfect example of an oscillating system because the child will be moving forwards and backwards, alternately reversing the direction of motion with time.

7 0

3 years ago

Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin

erastova [34]

Answer

(C).

When there is an angle between the two directions, the cosine of the angle must be considered.

Step by step Solution

The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.

The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.

The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is $W=F\times d$ .

Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the $sin(90)=0$ meaning that there is no work done in the direction perpendicular to the motion.

The third answer" When there is an angle between the two directions, the cosine of the angle must be considered." is correct because the work is calculated using the force in the direction of the motion. The magnitude of this force is $F\times d\times \cos(\theta).$

4 0

4 years ago

Read 2 more answers

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

3 years ago

_______ are different forms of a single element. A) Atoms B) Elements C) Ions D) Isotopes

ELEN [110]

The answer is D. Isotopes.
Hope that helped.

7 0

4 years ago

Read 2 more answers

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

4 years ago

What's an anime I should watch next?? *WEEPS ONLY* ¨ˆ¨

What's an anime I should watch next?? WEEPS ONLY ¨ˆ¨