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rosijanka [135]
3 years ago
14

How does a theory differ from a theory

Physics
2 answers:
mafiozo [28]3 years ago
8 0

Both of those terms are widely used in higher education, particularly in the sciences.  Their technical definitions, as well as their common underlying meanings, are precisely identical.  So you can use either one without any fear of being corrected or misunderstood.

Lyrx [107]3 years ago
3 0

Answer:

A scientific theory is a well-substantiated explanation of some aspect of the natural world, based on a body of facts that have been repeatedly confirmed through observation and experiment. Such fact-supported theories are not "guesses" but reliable accounts of the real world.

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Science assessment help pls
DENIUS [597]

Answer:

Work Done = W

force = F

Distance = d

W = Fd

or W = F*d

W (in joules) = 3.5*4 = 14 Nm (or J)

1Nm = 1J

so newton meters and joules are the same

Power = Work (in joules) /time (in seconds)

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3 0
3 years ago
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Two charges of +4x10-SC and -4x10-5C are placed 1 meter apart. Using Coulomb's law, the force between them is 14.4 N.
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Answer:

d. the force is reduced by one quarter to 3.6 n

Explanation:

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2 years ago
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Answer:

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3 years ago
3. Maverick and Goose are flying a training mission in their F-14. They are
Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

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When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

y=y_0 - \frac{gt^2}{2}    [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

x = v.t     [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

t=17.5 sec

Replacing in [2]

x = 688\ m/sec \ (17.5sec)

x = 12040\ m

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

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Answer:

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