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BARSIC [14]
3 years ago
9

A student completed a lab report. Which correctly describes the difference between the "Question" and "Hypothesis" sections

Physics
2 answers:
galben [10]3 years ago
4 0

Answer:

the second one!

Explanation:

the question is well, the question, a hypothesis is an educated guess on what you think will be the outcome

kramer3 years ago
4 0

Answer:

The Answer Is B. On Edge

Explanation:

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Luke stands on the edge of a roof throws a ball downward. It strikes the ground with 100J of kinetic energy. Luke now throws ano
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A. Less than 100 this the answer
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A real image can be obtained with:
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Answer:

convex lens and a concave mirror

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Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the ap
Verizon [17]

Answer:

The answer is "1.144 \times 10^{-4} \ m".

Explanation:

w=\frac{2.44 \lambda L}{D}\\\\D=\frac{2.44 \lambda L}{w}\\\\

   =\frac{2.44 \times 633 \times 10^{-9}\times 4 }{0.054}\\\\=\frac{6178.08\times 10^{-9}}{0.054}\\\\=1.144 \times 10^{-4} \ m

8 0
3 years ago
Select all that apply. The force that opposes the start of motion is referred to as _____.
77julia77 [94]
The correct answers are <span>starting friction and </span>static friction

Friction slows down all forces, but starting friction slows down or stops completely the start of motion.

8 0
3 years ago
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A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
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