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2 years ago

Which of the following cannot be determined from the spectrum of a star?

1 answer:

4 0

I think the correct answer from the choices listed above would be the last option. It is the chemicals in the core of the star that cannot be determined from the spectrum of a star. Spectrum shows the different classification of the stars depending on their spectral characteristics. It usually involves the light, the wavelength and the distance.

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A 2500-kg car is being pushed up a hill at an angle of 35 degrees. Determine the gravitational

velikii [3]

Answer:

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

Explanation:

For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.

sin 35 = Wₓ / W

cos 35 = W_y / W

Wₓ = W sin 35

W_y = W cos 35

Wₓ = 2500 9.8 sin 35

Wₓ = 14052.6 N

let's write the equations for each axis

and

Y axis

N-W_y = 0

N = W_y

X axis

F -Wₓ = m a

F = Wₓ + m a = mg sin 35 + m a

F = m (a + g sin 35)

let's calculate

F = 2500 (5 + 9.8 sin 35)

F = 26552.6 N

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

3 0

2 years ago

*example included* Two uncharged spheres are separated by 3.50 m. If 1.30 ✕ 10¹² electrons are removed from one sphere and place

likoan [24]

Considering the Coulomb's Law, the magnitude of the Coulomb force is 3.1865 N.

<h3>Coulomb's Law</h3>

Charged bodies experience a force of attraction or repulsion on approach.

From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.

From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=k\frac{Qq}{d^{2} }$

where:

F is the electrical force of attraction or repulsion. It is measured in Newtons (N).
Q and q are the values of the two point charges. They are measured in Coulombs (C).
d is the value of the distance that separates them. It is measured in meters (m).
K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹ $\frac{Nm^{2} }{C^{2} }$ .

The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

In this case, you know that:

The two uncharged sphere are separated by the distance of d= 3.50 m
The number of electrons are 1.30×10¹².
Electrons is elementary charge and charges on both the sphere is same. The value of electron is 1.602×10⁻¹⁹ C. This is, Q=q=1.30×10¹²×1.602×10⁻¹⁹ C= 2.0826×10⁻⁷ C

Replacing in Coulomb's Law:

$F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(2.0826x10^{-7} C)x(2.0826x10^{-7} C)}{(3.50 m)^{2} }$

Solving:

Finally, the magnitude of the Coulomb force is 3.1865 N.

Learn more about Coulomb's Law:

brainly.com/question/26892767

#SPJ1

7 0

10 months ago

Which statement is true?

miv72 [106K]

B might be the correct answer

8 0

2 years ago

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates

34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance $C_{0}$ = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed $3.00 \times 10^{4}$ V/m?