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kramer
3 years ago
5

3 objects that have gravitational potential energy?

Physics
2 answers:
WITCHER [35]3 years ago
8 0
When an object is above the Earth's surface it has gravitational potential energy (GPE). The amount of GPE an object has depends on its mass and its height above the Earth's surface.
The weight of an object is the size of the force of gravity pulling the object down. When an object falls it transfers GPE to KE (kinetic energy).
sleet_krkn [62]3 years ago
3 0
The water in the tank above the buildings
The fruits on the tree
Something like a cup or a book or a laptop on the table
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Un soldado de 1.80 m de altura ha sido herido por una bala, cuya masa es de 100 g. Deciden colocar al soldado en una centrifugad
maks197457 [2]

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3 years ago
11. Explain why in terms of gravity and air resistance and the 2nd law, objects in free fall regardless of mass hit at the same
Alchen [17]

Answer:

When an object is dropped, it accelerates toward the center of Earth. Newton's second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force.

7 0
3 years ago
Pls help me l will make it brainlest ​
nikklg [1K]

Answer:

<u>0.04 °C⁻¹</u>

Explanation:

First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.

<u />

=============================================================

Finding Linear Expansivity :

⇒ α = Final length - Original length / (Original length × ΔT)

⇒ α = 9 - 4 / (4 × 70 - 20)

⇒ α = 5 / 5 × 50

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Finding Area Expansivity :

⇒ Area Expansivity = 2 × Linear Expansivity

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3 0
1 year 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
3 years ago
A 1200.0-kg car is traveling at 19m/s. The driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic f
Alja [10]
<h2>Answer</h2>

option D)

2.4 seconds

<h2>Explanation</h2>

Given in the question,

mass of car = 1200kg

speed of car = 19m/s

Force due to direction of travel

F = ma

  = 12000(a)

Force to due frictional force in reverse direction

-F = mg(friction coefficient)

   = -12000(9.81)(0.8)

<h2>-mg(friction coefficient) = ma  </h2>

(cancelling mass from both side of equation)

g(0.8) = a

(9.81)(0.8) = a

a = 7.848 m/s²

<h2>Use Newton Law of motion</h2><h3>vf - vo = a • t</h3>

where vf = final velocity

          vo = initial velocity

          a = acceleration

           t = time

0 - 19 = 7.8(t)

t = 19/7.8

  = 2.436 s

  ≈ 2.4s

5 0
2 years ago
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