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beks73 [17]
4 years ago
8

¿ES CIERTO QUE PARTE DE LAS INTERFERENCIAS DE LA TELE SON RADIACIONES DEL BIG BANG

Physics
1 answer:
Artemon [7]4 years ago
6 0
Pienso que no puesto que el "Big Bang" es nada más una teoría, no es algo comprobado por los científicos. Así que parte de las interferencias de la tele provienen de otra parte. ‍♀️
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Why is the mass of carbon atom greater than the total mass of its protons and electrons
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One or more neutrons in the nucleus add mass to the atom.
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4 years ago
Can someone please help me?
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Explanation:

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4 years ago
A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 4.4 cm from the axis of rotation. (a) Calcul
romanna [79]

Answer:

a) a =0.53 m/s²

b) μ=0.054

c) μ = 0.068

Explanation:

a) If we assume that the turntable is rotating at a constant speed, the only force acting on the seed parallel to the surface, which keeps it  from following a straight line trajectory, is the centripetal force.

So, we can apply Newton's 2nd Law to the seed in this way:

Fnet = m*a = m*ac = m*ω²*r

We have the value of the angular speed, ω, in rev/min, so it is advisable to convert it to rad/sec, as follows:

ω = 33 rev/min*(1 min/60 sec)*(2*π rad/ 1 rev) = 11/10*π rad/sec

So, replacing in (1), we can solve for ac, as follows:

ac = ω²*r = (11/10)²*π²*0.044 m = 0.53 m/s²

b) Now, the centripetal force that we found above, is not a new type of force, it must be a force that explains the behavior of the seed.

As the seed does not slip, the only force acting  on it parallel to the surface, is the static  friction force, which has a maximum value, as follows:

Ff = μ*N

As there is no movement in the vertical direction, this means that the normal force must be equal and opposite to Fg, so we can write the expression for Ff as follows:

Ff = μ*m*g

Now, this force is no other than centripetal force, so we can write this equation:

Ff  = Fc ⇒ μ*m*g = m*ac

⇒ μ*g = ac

Solving for μ:

μ = ac/g = 0.53 m/s² / 9.8 m/s² = 0.054

c) During the acceleration period, added to the centripetal acceleration, as the angular speed is not constant, we will have also an angular acceleration, γ , which we can get as follows:

γ = Δω/Δt = (11/10)*π / 0.37 s = 9.34 rad/sec²

By definition of angular acceleration, there exists a fixed  relationship between the angular acceleration and the tangential acceleration (same as the one between angular and tangential speed), as follows:

at = γ*r = 9.34 rad/sec²*0.044 m = 0.41 m/s

When the turntable has reached to its maximum angular velocity, it will have also the maximum value of the centripetal acceleration, which we have just found out.

So, the magnitude of the total acceleration (at the moment of maximum acceleration) as they are perpendicular each  other) , is given by the following expression:

a = √(ac)²+(at)² = 0.67 m/s²

Now, as friction force opposes to the relative movement between surfaces (the seed and the turntable), it shall be larger than the product of the mass times the total acceleration, acting along  the same action line, so we can say:

Ffmin = μ*m*g = m*a

⇒ μmin = a/g = 0.67 m/s²/9.8 m/s² = 0.068

5 0
4 years ago
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3 years ago
A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically dr
yan [13]

Answer:

49.79 m/s

Explanation:

Given:

Initial velocity of the roller coaster is, u=0\ m/s

Vertical drop or the displacement of the roller coaster is, y=-126.5\ m

The displacement is negative as the motion is in downward direction.

Now, as the motion is in vertical direction only, the acceleration of the roller coaster will be due to gravity acting in the downward direction.

So, the acceleration of the roller coaster is, a=g=-9.8\ m/s^2

Now, using the following equation of motion:

v^2=u^2+2ay

Where, 'v' is the velocity of the roller coaster at the bottom.

Plug in all the given values and solve for 'v'. This gives,

v^2=0^2+2(-9.8)(-126.5)\\\\v^2=2479.4\\\\v=\sqrt{2479.4}\\\\v=49.79\ m/s

Therefore, the speed of the roller coaster at the bottom of the drop is 49.79 m/s.

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