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Andru [333]
3 years ago
13

what is the most likely reason that a hypothesis would be useful even if experiments show that is false

Physics
2 answers:
svetoff [14.1K]3 years ago
7 0

Answer:

An observation is made. To understand why the phenomenon is occurring, a hypothesis is built up which can be tested by a scientific method. The result of the experiment can be true or false.

If the result false, it implies that the dependent and independent variables are not related and a new experiment must be designed or a modification is required in the experiment.

Thus, even if the result is false, the hypothesis would be useful because conclusions can be drawn from it and next course of action can be defined.

Vinil7 [7]3 years ago
4 0
You are able to test your hypothesis if you do your experiment again.
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grigory [225]

Answer:

PEENNISS

Explanation:

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Volcanic ash and sulfur dioxide spewed out of Mt. Pinatubo in 1991. These materials can reflect incoming solar radiation. Over t
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The suspended ash made for some some spectacular sunsets! Sulfuric acid was spread worldwide, increasing acidity of rain. Ash deflected energy from the sun, causing a slight drop on global temps for a few years.
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A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25
solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

k=\frac {F}{x}

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

k=\frac {80}{0.25}=320N/m

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3 0
4 years ago
In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

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3 years ago
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