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3 years ago

Which is the dependent variable?

Physics

2 answers:

Bess [88]3 years ago

5 0

Answer:

Number of stars

Explanation:

The given figure shows a graph between the number of stars on the y-axis and constellations on the x-axis. The independent variable is placed on the x-axis and dependent variable on the y-axis. Dependent variable is the one whose value changes when the independent variable is changed. As the constellations vary, the number of stars vary. Thus, number of stars is dependent variable.

zmey [24]3 years ago

4 0

Number of Stars ( Brainliest) if you dont care :)

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The figure below shows a man in a boat on a lake. The man's mass is 74 kg, and the boat's is 135 kg. The man and boat are initia

vazorg [7]

The velocity of the boat after the package is thrown is 0.36 m/s.

<h3>Final velocity of the boat</h3>

Apply the principle of conservation of linear momentum;

Pi = Pf

where;

Pi is initial momentum
Pf is final momentum

v(74 + 135) = 15 x 5

v(209) = 75

v = 75/209

v = 0.36 m/s

Thus, the velocity of the boat after the package is thrown is 0.36 m/s.

Learn more about velocity here: brainly.com/question/6504879

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2 years ago

Two tractors pull against a 1000 kg log. If the angle of the tractors' chains in relation to each other is 18.0° (each 9.0° from

Morgarella [4.7K]

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4 years ago

Find the magnitude of the electric force between the charges 0.12 C and 0.33 C at a separation of 2.5 m. k=8.99×109N⋅m2/C2.

makkiz [27]

Answer:

F= k Q1 Q2 / r^2

plug in the numbers

8 0

3 years ago

What is the gravitational potential energy of a 3 kg ball kicked into the air at a height of 5 meters?

sladkih [1.3K]

formula for gravitational P.E =mgh

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3 years ago

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