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Arte-miy333 [17]
3 years ago
8

When a hypothesis is proven wrong, scientists often begin by revising the . After that, it’s common for scientists to redesign t

he .
Physics
2 answers:
jasenka [17]3 years ago
6 0

Answer:    Hypothesis, experiment

Explanation:  When scientists set up a hypothesis, which is an assumption that needs to be proven by an experiment, they conduct an experiment based on starting assumptions. If the experiment shows that the hypothesis is incorrect, then they have to revise the hypothesis. On the basis of the revised hypothesis, scientists are setting a new experiment to prove the new hypothesis as true. So hypotheses are revised and new experiments are put in place until an experiment proving the revised hypothesis as accurate, which is no longer a hypothesis, can be considered as scientific theory. Everything that is empirically proven through an experiment is considered true.

Rainbow [258]3 years ago
3 0
1) thesis
2) experiment
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Two point charges lie on the x axis. A charge of + 2.30 pC is at the origin, and a charge of − 4.50 pC is at x=−11.0cm.
Radda [10]

r₁ = distance of point A from charge q₁ = 0.13 m

r₂ = distance of point A from charge q₂ = 0.24 m

r₃ = distance of point A from charge q₃ = 0.13 m

Electric field by charge q₁ at A is given as

E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C     towards right

Electric field by charge q₂ at A is given as

E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C    towards left

Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction

Electric field at A will be zero when

E₁ = E₂ + E₃

1.225 = 0.703 + E₃

E₃ = 0.522 N/C

Electric field by charge "q₃" is given as

E₃ = k q₃ /r₃²

0.522 = (9 x 10⁹) q₃/(0.13)²

q₃ = 0.980 x 10⁻¹² C = 0.980 pC

4 0
3 years ago
A toy train engine rests motionless on a track. One student begins pushing the engine to the right with a force of 2 newtons. At
Svetradugi [14.3K]
I think it’s B

Hope it helps
8 0
3 years ago
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0
3 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
Can you answer any of these?
givi [52]
#39. rain, snow, sleet and hail.
3 0
3 years ago
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