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Anuta_ua [19.1K]
3 years ago
9

What is the "basics" of a proper experiment?

Physics
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer:

sorry in my sense, an experiment once only changes one variable and need a control setup for experimental setup to make sure is fair test

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Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b
Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

\theta = Angle of slide = 20.7°

\mu = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J

The work done by the force of gravity is 23.52092 J

8 0
3 years ago
This property of waves is the only property where the relationship between energy and this property are indirect or inverse
yaroslaw [1]

Answer: I don't understand

Explanation:

study and pay attention

4 0
3 years ago
How is energy harnessed to do useful work?
baherus [9]

Answer:

But there are ways to harness kinetic energy to either generate useful mechanical work or electricity. This is what many have tried to do to make use of energy that would be otherwise wasted. One way to harness kinetic energy that has popped up many times in recent years has to do with roads and speed bumps

Explanation:

8 0
3 years ago
A client has developed dystrophic calcification as a result of macroscopic deposition of calcium salts. The tissue that would be
Komok [63]

Answer:

Tissues that are damaged or injured.

Explanation:

Dystrophic calcification involves the deposition of calcium in soft tissues despite no disturbance in the calcium metabolism, and this is often seen at damaged tissues.

Examples of areas in the body where dystrophic calcification can occur include atherosclerotic plaques and damaged heart valves.

4 0
3 years ago
Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o
adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

F = \sqrt{F_h^2+F_v^2}

F = \sqrt{3.36^2+13.98^2}

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

\theta =tan^{-1}(\dfrac{F_v}{F_h})

\theta =tan^{-1}(\dfrac{13.98}{3.36})

   θ = 76.48°

3 0
3 years ago
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