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DerKrebs [107]
3 years ago
10

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

#SPJ1

7 0
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]
Pie ?????????????? π pizza ??? wut
7 0
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

Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²

7 0
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
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