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stich3 [128]
3 years ago
6

which of following categories of small bodies in the solar system does ceres belong to?. . . A.Dwarf Planets. B.Comets. C.Astero

ids. D.Planets
Physics
2 answers:
FromTheMoon [43]3 years ago
6 0
I suggest you to choose A. as ceres belong to Dwarf Planets.
olchik [2.2K]3 years ago
5 0
The category of small bodies that the solar system Ceres belongs to is the Dwarf Planets. The correct answer is A, dwarf planets.
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Question 2
Ksivusya [100]

5.77 ×10^1^4 Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

<h3>Given:</h3>

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

            c = Speed of light = 3 × 10^8 m/s

            λ = Wavelength of light

∴ f = c / λ

  f = \frac{3*10^8}{520 * 10^-^9}

    = 5.77 ×10^1^4 Hz

Therefore,  5.77 ×10^1^4 Hz is the green photon's frequency .

Learn more about wavelength here:

brainly.com/question/10728818

#SPJ1

4 0
2 years ago
Which of the following is equivalent to 4.5 m?
Romashka [77]
We can’t see the following
3 0
3 years ago
ITS TIMED PLEASE HELP !!!!!
dusya [7]

Answer: direct linear

Explanation:

3 0
3 years ago
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
Given the function f(x) = 8x3 − 3x2 − 5x 8, what part of the function indicates that the left and right ends point in opposite d
ale4655 [162]

Answer: The degree of the first term.

Explanation:

The function:

f(x) = 8x^3-3x^2-5x^8

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.

Let the left hand side be donated by -x.

Then,

f(-x) =8(-x)^3-3(-x)^2-5(-x)^8\\ \Rightarrow f(-x)=-8x^3-3x^2-5x^8

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.

8 0
4 years ago
Read 2 more answers
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