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katen-ka-za [31]
3 years ago
15

The time between two successive meridian crossings of the vernal equinox (or any other fixed point in the celestial sphere) is c

alled ___________ day and it is ___________ long.
Physics
1 answer:
Ludmilka [50]3 years ago
8 0

The best answers to complete this sentence would be the following:

 

“sidereal day”

“upper median”

 

<span>It keeps the time, in the two successive upper meridian of the sun during the crossing. Also it is when the star crosses the celestial meridian.</span>

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An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro
svet-max [94.6K]
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
6 0
2 years ago
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

 

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

         a = 1.1 10⁵ m / s²

6 0
2 years ago
Un móvil se desplaza con una rapidez inicial de 100 km/h, se le aplican los frenos con una
KIM [24]

Answer:

in English please I am quite puzzled

4 0
3 years ago
According to newton's second law of motion, how much force will be required to accelerate an object at the same rate if its mass
guajiro [1.7K]

<u>D.</u>

For every action ther is an equal and opposite reaction.

equal in force, opposite in direction

5 0
3 years ago
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s
IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

  = 149.668 * 10^8 m

next we will express the distance between the earth and the sun

r = \frac{a(1-E^2)}{1+Ecos\beta }   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

\beta = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

v^2 = \frac{4\pi^2 }{r_{c} }   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

(V^2 = (\frac{4\pi^{2}  }{149.626*10^9})

therefore : V = 1.624* 10^-5 m/s

4 0
3 years ago
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