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3 years ago

The best scientific reason for a scientist to accept a specific theory is?

Physics

1 answer:

lions [1.4K]3 years ago

7 0

Answer:

The theory is supported by all the available observations and data.

Explanation: The scientific community will accept a theory when a sufficient body of evidence supports it. This includes experiments that refute other potential theories. Experiments should also be carried out that attempt to disprove the theory but cannot.

It should not matter who proposed the theory or who supports it, and instead should entirely be based on the quality and abundance of data supporting it.

Hope this helps!

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-ya girl Lizzie

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If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the

GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height = -10 m

$y_{0}$ : is the initial height = 0

$v_{0_{y}}$ : is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

$t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s$

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

$x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}$

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

$x_{f} = 5.51 m/s*1.43 s = 7.88 m$

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

4 0

3 years ago

In any electromagnetic wave,

KATRIN_1 [288]

C.half the energy is carried by the electric field and half is carried by the magnetic field.

3 0

3 years ago

A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry

Sauron [17]

Answer:

a) E = -4 10² N / C , b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

$F_{e}$ - $T_{x}$ = m a

Axis y

$T_{y}$ - W = 0

Initially the system is in equilibrium, so zero acceleration

Fe = $T_{x}$

T_{y} = W

Let us search with trigonometry the components of the tendency

cos θ = T_{y} / T

sin θ = $T_{x}$ / T

T_{y} = cos θ

$T_{x}$ = T sin θ

We replace

q E = T sin θ

mg = T cosθ

a) the electric force is

$F_{e}$ = q E

E = $F_{e}$ / q

E = -0.032 / 80 10⁻⁶

E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

q E / mg = tan θ

θ = tan⁻¹ qE / mg

Let's calculate

θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

θ = tan⁻¹ 0.3265

θ = 18 ⁰

sin 18 = x/0.30

x =0.30 sin 18

x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

$F_{e}$ = m aₓ

aₓ = q E / m

aₓ = 80 10⁻⁶ 4 10² / 0.01

aₓ = 3.2 m / s²

Axis y

W = m $a_{y}$

a_{y} = g

a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

a = √ aₓ² + a_{y}²

a = √ 3.2² + 9.8²

a = 10.31 m / s²

The Angle meet him with trigonometry

tan θ = a_{y} / aₓ

θ = tan⁻¹ a_{y} / aₓ

θ = tan⁻¹ (-9.8) / 3.2

θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

8 0

3 years ago

A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t

Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0

3 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

3 years ago