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Degger [83]
3 years ago
8

1. Name several constellations that are visible every night here in Bellingham, WA during all 12 months of the year. Why are the

se constellations always visible whereas others are seasonal?
Physics
1 answer:
Leya [2.2K]3 years ago
8 0

Answer:

Big Dipper, Little Dipper, Cassiopeia, Cepheus

Explanation:

In this region, the above constellations are circumpolar. This means that they appear above the horizon at all times. These are only visible all rear round for people living in Canada and Northern United States.

Circumpolar constellations are constellations that never appear below the horizon when seen from a particular location on planet Earth. Furthermore, these constellations can be seen all year while others are only seen at specific times during the year; thus they are known as seasonal constellations.

Five northern constellations are visible from most locations that are north of the equator. These are Cassiopeia, Cepheus, Draco, Ursa Major, and Ursa Minor.

Note that  Ursa Major is often confused with the Big Dipper. While the Little Dipper (which is much fainter) is found in the Ursa Minor constellation. Cassiopeia can be recognized due to its W shape which is quite prominent.  

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Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma
Helga [31]

Answer:

<em>Which box has the greater acceleration?</em>

E. Box A

Explanation:

<em>The question is incomplete:</em>

<em>Which box has the greater acceleration?</em>

The bouyant force exerted by the water is equal in both boxes, because it depends on the volume displaced (that is the same for both boxes) and the density of the water.

But, the weight of each boxes is different, according to their density.

For the Box A the acceleration will be:

m_aa_a=gV(\rho_w-\rho_a)\\\\\rho_aVa_a=gV(\rho_w-\rho_a)\\\\a_a=g\frac{(\rho_w-\rho_a)}{\rho_a}

The same applies for the Box B:

a_b=g\frac{(\rho_w-\rho_b)}{\rho_b}

If we express the ratio of the accelerations, we have:

a_a/a_b=\frac{(\rho_w-\rho_a)}{\rho_a}*\frac{\rho_b}{(\rho_w-\rho_b)}\\\\

a_a/a_b=\frac{(\rho_w-\rho_a)}{(\rho_w-\rho_b)} \frac{\rho_b}{\rho_a}

We know that both densities are lower than water, because they accelerate upward to the surface when they are released (if they were more dense than water, they would sink more).

We will treat the densities as relative to water, so it becomes rho_w=1.

If we distribute the product, and know that the density of B is higher than the density of A, and both are higher than the product of the densities, we have:

\rho_w=1\\\\\frac{a_a}{a_b}=\frac{(1-\rho_a)}{(1-\rho_b)} \frac{\rho_b}{\rho_a}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}\\\\\\\rho_b>\rho_a>\rho_a\rho_b>0\\\\\\\frac{a_a}{a_b}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}>1\\\\a_a>a_b

The acceleration of A is higher than the acceleration of B.

7 0
4 years ago
If the sum of the external forces on an object is zero, then the sum of the external torques on it
Zarrin [17]

Answer:

True.

Explanation:

If the sum of the external forces on an object is zero, then the sum of the external torques on it  must also be zero.

The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.

Hence, the given statement is true.

5 0
3 years ago
Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.
nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

  • The kinetic energy ratio is   \frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2
<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

          K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

  • Translational kinetic energy. Due to the linear movement of the body

            K_{tras} =\frac{1}{2} m v^2

  • Rotational kinetic energy. Due to the rotational movement of the body.

            K_{rot} = \frac{1}{2} I w^2

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

           K_{earth} = K_{tras} + K_{rot}

Linear and rotational speed are related.

           v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

           I = \frac{2}{5 }  m r^2  

Let's  subatitute.

         

          K_{earth} = \frac{1}{2} \  m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}  

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

       w = \frac{2 \pi}{T}  

Let's substitute.

        K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2  \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2  

        K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)  

Data for Earth are tabulated:

  • Mass m = 5.98 1024 kg
  • Radius r = 6.37 10⁶ m
  • Radius orbits tras = 1.496 10¹¹ m
  • Rotation period T_{rot} = 24 h (\frac{3600s}{1h}) = 8.64 10⁴s
  • Translation period  T_{tras} = 365 d (\frac{24h}{1 d}) (\frac{3600s}{1h}) = 3.15 10⁷ s

Let's calculate.

        K_{earth} = 4 \pi^2 5.98 \ 10^{24}  ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2  \ +  \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )

        K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)

        K_{earth}= 2.66 \ 10^{33} J

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of T_{Sum} = 27 days.

The kinetic energy of the sun is;

          K_{sum} = K_{rot} =  \frac{1}{2} I w^2  

          K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2  

          K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2  

The tabulated data for the sun are:

  • Mass m = 1,991 1030 kg.
  • Radius R = 6.96 10⁸ m
  • Period T = 27 d (\frac{24h}{1 d} ) (\frac{3600s}{1h}) = 2.33 10⁶ s

         

Let's calculate.

           

          K_{sum} = 1.40 \ 10^{36} J

The relationship of the kinetic energy of the sun and the Earth is:

        \frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}  

       \frac{K_{sum}}{K_{earth}} =  5.3 \ 10^2  

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

  • The kinetic energy ratio is:  \frac{K_{Sum}}{K_{Earth}} = 5 \ 10^2

Learn more about kinetic energy here: brainly.com/question/25959744

5 0
3 years ago
he deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudden
Lubov Fominskaja [6]

Answer:

Incomplete question

This is the complete question

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 V potential difference is suddenly applied to the initially uncharged plates through a 1000 Ω resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95 V?

Explanation:

Given that,

The dimension of 10cm by 2cm

0.1m by 0.02m

Then, the area is Lenght × breadth

Area=0.1×0.02=0.002m²

The distance between the plate is d=1mm=0.001m

Then,

The capacitance of a capacitor is given as

C=εoA/d

Where

εo is constant and has a value of

εo= 8.854 × 10−12 C²/Nm²

C= 8.854E-12×0.002/0.001

C=17.7×10^-12

C=17.7 pF

Value of resistor resistance is 1000ohms

Voltage applied is V = 100V

This Is a series resistor and capacitor (RC ) circuit

In an RC circuit, voltage is given as

Charging system

V=Vo[1 - exp(-t/RC)]

At, t=0, V=100V

Therefore, Vo=100V

We want to know the time, the voltage will deflect 95V.

Then applying our parameters

V=Vo[1 - exp(-t/RC)]

95=100[1-exp(-t/1000×17.7×10^-12)]

95/100=1-exp(-t/17.7×10^-9)

0.95=1-exp(-t/17.7×10^-9)

0.95 - 1 = -exp(-t/17.7×10^-9)

-0.05=-exp(-t/17.7×10^-9)

Divide both side by -1

0.05=exp(-t/17.7×10^-9)

Take In of both sides

In(0.05)=-t/17.7×10^-9

-2.996=-t/17.7×10^-9

-2.996×17.7×10^-9=-t

-t=-53.02×10^-9

Divide both side by -1

t= 53.02×10^-9s

t=53.02 ns

The time to deflect 95V is 53.02nanoseconds

5 0
3 years ago
Which parts of the roller coaster determine the amount of potential energy in the system? Explain your answer.
Paha777 [63]

Answer:

the lowest point, where the amount of energy is the greatest

Explanation:

The correct answer is "The roller coaster cars determine the amount of kinetic energy in the system. It’s the only part of the system that moves. Kinetic energy depends on the velocity and mass of a moving object."

6 0
3 years ago
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