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d1i1m1o1n [39]
3 years ago
14

At what point does the comet experience the strongest force of gravity?

Physics
1 answer:
spin [16.1K]3 years ago
5 0
I think the answer may be the letter B
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A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle
dangina [55]

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

\phi = BAcos\theta where:

\phi is the magnetic flux

B is the magnetic field

A is the cross sectional area

\theta is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

\phi = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>

4 0
3 years ago
Heres a random question just for fun and out. of boredom ok so who started the nasa program? and why? what made him want to do t
Pachacha [2.7K]
Eisenhower started the NASA project to develop technology for military application.
5 0
3 years ago
How are electromagnetic waves different from all other waves
Verdich [7]

Answer:

Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.

Explanation:

6 0
4 years ago
Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
jasenka [17]

Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
3 years ago
If a negatively charged rod is held near a neutral metal ball, the ball is attracted to the rod. this happens because
notsponge [240]
The can be found elsewhere and as follows:


<span>A. of magnetic effects.
B. the ball tries to pull the rod’s electrons over to it.
C. the rod polarizes the metal.
D. the rod and the ball have opposite charges. 
</span><span>
I believe the correct answer is option C. If a negatively charged rod is held near a neutral metal ball, the ball is attracted to the rod. this happens because </span>the rod polarizes the metal. Hope this answers the question.
5 0
4 years ago
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