What is meant by saying that a hypothesis must be testable ?

In electrical work, silver is used as a(n)

dmitriy555 [2]

Answer: [A]: conductor .
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7 0

3 years ago

A uniform, solid disk with mass m and radius R is pivoted about a horizontal axis through its center. A small object of the same

Alekssandra [29.7K]

Answer:

$\omega=\sqrt{\frac{4g}{3R}}$

Explanation:

According to the law of the conservation of energy, the gravitational potential energy of the physical system will be converted into rotational kinetic energy. So, we have:

$U=K_{rot}\\mgR=\frac{I_s\omega^2}{2}$

$I_s$ Is the moment of inertia of the system, that is the sum of the moments of inertia of the disk and the small object.

$mgR=\frac{(I_d+I_o)\omega^2}{2}\\mgR=\frac{(\frac{mR^2}{2}+mR^2)\omega^2}{2}\\mgR=\frac{(\frac{3}{2}mR^2)\omega^2}{2}\\gR=\frac{3}{4}R^2 \omega^2\\\omega^2=\frac{4g}{3R}\\\omega=\sqrt{\frac{4g}{3R}}$

4 0

3 years ago

The sun emits energy in the form of electromagnetic waves produced by nuclear reactions deep in the sun's interior. Assume that

tatiyna

Answer:

I = W / 4π R_{s}², P = W / 2π c R_{s}², Io /I_{earth} = 10⁴

Explanation:

The intensity is defined as the ratio between the emitted power and the area of the spherical surface

I = P / A

Since the emitted power is constant and has a value of W for this case, let's look for the area of the sphere on the surface of the sun

A = 4π $R_{s}$ ²

I = W / 4π R_{s}²

.- The radiation pressure for total absorption is

P = S / c

Where S is the Pointer vector that is equal to the intensity

Let's replace

P = W / 2π c R_{s}²

.- We repeat for r = R_{s}/2

I₂ = W / 4π (R_{s}/ 2)²

I₂ = 4 W / 4π R_{s}²

I₂ = 4 Io

I₀ = W / 4piRs2

We calculate the radiation pressure

P₂ = I₂ / c

P₂ = 4 I₀ / c

P₂ = 4 (W / 4pi c Rs2)

.- the relationship between these magnitudes is

I₂ / I₀ = 4

P₂ / P₀ = 4

Let's calculate the intensity on the surface where the Earth is

r = 1.50 10¹¹ m

$I_{earth}$ = W / 4π r²

Io / I_{earth} = r² / $R_{s}$ ²

Io /I_{earth} = (1.5 10¹¹ / 6.96 10⁸) 2

Io /I_{earth} = 4.6 10⁴

Io /I_{earth} = 10⁴

4 0

4 years ago

show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t

timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

(c) ratio = 1800

Explanation: (a) <u>Potential</u> <u>difference</u> is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

so

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is $-109.44$ x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$