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lorasvet [3.4K]
4 years ago
6

An example of a Natural electric field is

Physics
2 answers:
liberstina [14]4 years ago
6 0

Answer:

Lightning

Explanation

user100 [1]4 years ago
5 0
Possibly an eel
I’m not too sure ... but definitely lightning
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Which statement best describes the relationship between mass and gravitational attraction (pull)?
Gnesinka [82]

Answer:

The first option, "The more mass an object has, the greater the gravitational pull. "

Explanation:

Newton's Law of Gravity states that F_G=\frac{Gm_1m_2}{r^{2}}, where G is the gravitational constant, m_1 and m_2 are the masses of the two objects, and <em>r</em> is the distance between the objects' centers. Because the objects' masses are in the fraction's numerator, they are directly proportional to F_G, and increasing the mass of one or both objects increases the gravitational pull.

Please have a great day! I hope this helps you understand the question!

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3 years ago
A satellite of mass ????=238 kg is in circular orbit around the Earth at an altitude above the earth's surface equal to the eart
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Answer:

Explanation:

a) the speed=

V=√2gr²/2r

=√gr/2

√10*637/2

V=31850m/s²

Magnitude of the acceleration

a=Gm/r²

10*238/(6370)²

2380/12740

=0.187

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What is the difference between transformation of energy and transfer of energy​
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In energy transfer, happens from one object to another without changing its form. For example, when you fill a pot of water and sit it on the stove, the water continues to heat until you turn it off. While in a energy transformation, the energy changes forms such as Chemical, kinetic.

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Juno created a diagram to describe characteristics of freezing and deposition. Which labels belong in the areas marked X and Y?
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A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction
finlep [7]

consider the motion in x-direction

v_{ox} = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

a_{x} = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = v_{ox} t + (0.5) a_{x} t²

100 =  v_{ox} (4.60)

v_{ox} = 21.7 m/s


consider the motion along y-direction

v_{oy} = initial velocity in y-direction = ?

Y = vertical displacement  = 0 m

a_{y} = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = v_{oy} t + (0.5) a_{y} t²

0 = v_{oy} (4.60) + (0.5) (- 9.8) (4.60)²

v_{oy} = 22.54 m/s

initial velocity is given as

v_{o} = sqrt((v_{ox})² + (v_{oy})²)

v_{o} = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

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4 years ago
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