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3 years ago

Which of the following hypotheses is both falsifiable and testable?

2 answers:

5 0

B. If all plants are watered at least once a day then they will all grow at least 1 cm a day.

All plants grow at different rates.

Hope this helps! XD

Vladimir [108]3 years ago

5 0

Answer:

The correct answer to the question: Which of the following hypotheses is both falsifiable and testable:____, would be, B: If all plants are watered at least once a day then they will all grow at least 1 cm a day.

Explanation:

In order to carry out research, hypothesis must be generated that will ask the questions necessary to find out information on different topics. However, not all questions, or statements will serve as a source for research, or lead to research. For this to happen, a researched must know that there are conditions to how to formulate a hypothesis; the first condition is that the hypothesis must be verifiable, or testable, which means, it can be proven. The second is, that the hypothesis must be falsifiable, or refutable, by research, and the third, the hypothesis must lead to results that can lead to further testing and reproduction. In this case, saying that all plants, when watered once a day at least, will grow a minimum of 1 cm a day cannot be used because it is based on a wrong assumption, given that all plants grow at different rates, and not all can be watered in the same way. This, in itself, from the outset, invalidates the hypothesis.

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The spreading of waves behind an aperture is more for long wavelengths and less for short wavelengths.Less for long wavelengths

Allisa [31]

Answer:

Increase in wavelength of incident wave also increases the spread angle or spread of the interference pattern.

Explanation:

Solution:-

- The diffraction occurs when light bends in the same medium. The bending is the result of light waves "squeezing" through small openings or "curving" around sharp edges.

- Moreover, waves diffract best when the size of the diffraction opening (or grting or groove) corresponds to the size of the wavelength. Hence, light diffracts more through small openings than through larger openings.

- The formula for diffraction shows a direct relationship between the angle of diffraction (theta) and wavelength:

d sin (θ) = m λ

Where,

λ : Wavelength , θ : The spread angle , d : Slit opening or grating

- We can see that the wavelength λ and spread angle θ are related proportionally. So if we increase the wavelength of incident wave we also increase the spread angle or spread of the interference pattern.

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3 years ago

a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical

nlexa [21]

Answer:

Vx = 10.9 m/s , Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s × cos 55°

Vx = 10.9 m/s

Vx = V sin θ

Vy = 19 m/s × sin 55°

Vy = 15.6 m/s

6 0

3 years ago

Read 2 more answers

umka2103 [35]

You can hear a difference between these two sounds. That is because their pitch isdifferent. Pitch depends on the frequency of a sound wave. ... High sounds have highfrequencies and low sounds have lowfrequencies.

4 0

4 years ago

Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for

klemol [59]

The value of parameter C for the function in the figure is 2.

<h3>What is amplitude of a wave?</h3>

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(x - C)

where;

A is amplitude of the wave
C is phase difference of the wave

<h3>What is angular frequency of a wave?</h3>

Angular frequency is the angular displacement of any element of the wave per unit time.

From the blue colored graph; at y = 1, x = -2 cm

1 = cos(2 - C)

(2 - C) = cos^(1)

(2 - C) = 0

C = 2

Thus, the value of parameter C for the function in the figure is 2.

Learn more about phase angle here: brainly.com/question/16222725

#SPJ1

5 0

2 years ago

If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a

PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s $w_o$ = 12 rad/s

Angular Velocity at time 0.15s $w_f$ = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( $w_f$ - $w_o$ ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0

3 years ago