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Crank
3 years ago
10

Explain why a scientific theory cannot became a scientific law

Physics
1 answer:
Maslowich3 years ago
3 0

A common misconception is that scientific theories are rudimentary ideas that will eventually graduate into scientific laws when enough data and evidence have been accumulated. A theory does not change into a scientific law with the accumulation of new or better evidence.

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Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
Quick please and will give Brainliest!!!
masha68 [24]

25 nC

That is the answer

3 0
3 years ago
Read 2 more answers
Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons
yarga [219]

Answer:

q\approx 6.6\cdot 10^{13}~electrons

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

\displaystyle F=k\frac{q^2}{d^2}

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

\displaystyle q=\sqrt{\frac{F}{k}}\cdot d

Substituting values:

\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1

q = 1.05\cdot 10^{-5}~c

This charge corresponds to a number of electrons given by the elementary charge of the electron:

q_e=1.6 \cdot 10^{-19}~c

Thus, the charge of any of the spheres is:

\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}

\mathbf{q\approx 6.6\cdot 10^{13}~electrons}

5 0
2 years ago
A particle accelerator fires a proton into a region with a magnetic field that points in the +x-direction (a) If the proton is m
pochemuha

Answer:

a) -z direction, b) +z direction, c) F=0  

Explanation:

The magnetic force is given by the expression

        F = q v x B

the bold indicate vectors, this equation can be separated in its module

        F = a v B sin θ

and where θ is the angles between the speed and the magnetic field.

The direction of the force can be found with the right-hand rule. For a positive charge, the thumb goes in the direction of speed, the fingers extended in the direction of the magnetic field and the palm points in the direction of the force, if the charge is negative the force is in the opposite direction.

a) Let's apply this to our case

the proton is positively charged

moves in the direction of + x

The magnetized field goes in the direction of y

therefore applying the right hand rule the force must be in the direction of the negative part of the z-axis (-z)

The right-hand rule is used to find this address.

b)  in this case it indicates that the proton moves in the recode of -y

again we apply the right hand rule and the force is in the direction of + z

c)   The proton moves in the x direction

In this case the force is zero because the angle between the field and the speed is zero and the sine is zero, therefore the force is zero

4 0
3 years ago
Consider the hydrogen atom as described by the Bohr model. The nucleus of the hydrogen atom is a single proton. The electron rot
riadik2000 [5.3K]

To solve this problem we will apply the concept of centripetal acceleration. This type of acceleration is described as the product between the square of the angular velocity and the turning radius. Mathematically the expression can be expressed as

a_c = \omega^2 r

Here,

\omega =Angular velocity

r = Radius

Our values are given as,

\omega = 4.12*10^{16}rad/s

r = 5.29*10^{-11}

Replacing,

a_c = (4.12*10^{16})^2( 5.29*10^{-11})

a_c = 8.979*10^{22}m/s^2

Therefore the electron's centripetal acceleration is 8.979*10^{22}m/s^2

7 0
3 years ago
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