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IrinaK [193]
3 years ago
9

Which of the following statements about comets is true? A. Comets rarely fall into an orbit around the Sun. They usually enter t

he inner solar system only once and then are gone forever. B. Most comets have a circular orbit that keeps them in the space between Mars and Jupiter. C. Comets have very elliptical orbits that usually take them far beyond the orbit of Pluto, but also take them closer to the Sun than Earth. D. Comets have very elliptical orbits that usually take them closer to the Sun than Earth, but rarely do they get further away than Pluto.
Physics
2 answers:
Lelechka [254]3 years ago
7 0

the correct answer is C

GREYUIT [131]3 years ago
4 0
<h2>Right answer: Comets have very elliptical orbits that usually take them far beyond the orbit of Pluto, but also take them closer to the Sun than Earth</h2>

Comets are celestial bodies constituted by ice, dust and rocks that orbit around the Sun, after having been altered by the Oort cloud; following different trajectories that can be <u>highly eccentric elliptical</u><u> </u>(periodic trajectories), parabolic or hyperbolic.

One of the main characteristics of a comet is that it travels quite fast, on its way around the Sun and has a long tail. It should be noted that the tails of comets always go in the opposite direction to the Sun (due to the radiation pressure of sunlight).

Therefore, the correct option is C.

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An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce
Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Frequency of pendulum is given by

f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

Given in the question

f_p=\dfrac{1}{2}f_s

\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m

The unstretched length of the spring is 2.67 m

6 0
3 years ago
A water bath in a physical chemistry lab is 1.55 m long, 0.710 m wide, and 0.570 m deep (high). If it is filled to within 3.55 i
Lesechka [4]

Answer:

528 liter.

Explanation:

Volume of the tank(cuboid) = l*b*h

But volume of the water = l*b*h

Where

l= length of the tank

b = width of the tank

h = the length from the bottom of the tank,

3.55 in to m,

0.09017m

Length of the water in the tank = 0.570 - 0.09017

= 0.47983 m.

Volume = 0.47983*0.710*1.55

= 0.528 m3.

1 m3 = 1000 liter.

0.528 m3 = 0.528*1000

= 528 liter

7 0
3 years ago
30 km/h is _________m/s?<br> A.)8.3<br> B.)5.6<br> C.)13.9<br> D.)11.1
Juli2301 [7.4K]

Answer:

Correct  Option :-  A

Explanation:

3 0
2 years ago
) what is kinetic energy, and how does it differ from potential energy?
tankabanditka [31]
 Kinetic energy<span>is the </span>energy<span> of body or a system with respect to the motion of the body or of the particles in the system. </span>Potential energy<span> is the stored </span>energy<span> in an object of system because of its position or configuration.</span>
7 0
3 years ago
Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre
castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)  

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619  

=

t

+

C

Here C =constant of integration.   

x

=

0  at  t

=

0

, we get:  C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

(c)  required time for the jogger to run 6 miles is obtained by setting  

        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

7

/

10=

t

−

4.7619

         or

         t

=

0.832

h

r

s

6 0
3 years ago
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