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andrezito [222]
3 years ago
5

¿Qué diferencia una magnitud fundamental de una derivada?

Physics
2 answers:
artcher [175]3 years ago
7 0

Answer:

Explanation:

Una magnitud fundamental es aquella que se define por si misma y es independiente de las demás (masa, tiempo, longitud, etc.). magnitud derivada. Una magnitud derivada es aquella que se obtiene mediante expresiones matemáticas a partir de las magnitudes fundamentales (densidad, superficie, velocidad).

Alla [95]3 years ago
7 0
Una magnitud fundamental es aquella que se define por si misma y es independiente de las demás (masa, tiempo, longitud, etc.). magnitud derivada. Una magnitud derivada es aquella que se obtiene mediante expresiones matemáticas a partir de las magnitudes fundamentales (densidad, superficie, velocidad).


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Wil-E-Coyote drops a bowling ball off a cliff to try to catch the Roadrunner. The cliff is
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Answer:

t = 5.19 s

Explanation:

We have,

Height of the cliff is 132 m

It is required to find the time taken by the ball to fall to the ground. Let t is the time taken. So, using equation of kinematics as :

y=ut+\dfrac{1}{2}gt^2\\\\\text{since}\ u=0\\\\y=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2y}{g}}\\\\t=\sqrt{\dfrac{2\times 132}{9.8}}\\\\t=5.19\ s

So, it will take 5.19 seconds to fall to the ground.

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3 years ago
Which state of matter is the least compressible
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Incompressible. Compressibility is determine by the amount of space between particles in each state.  
7 0
3 years ago
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A motorist travels at 270 km in p hours at a certain average speed. an expression for the distance that this motorist travelled
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Average speed = total distance / time ⇒ total distance = average speed *  time

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5 0
3 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
How do magnetic trains work and what advantages do they have compared to regular trains
ad-work [718]

Answer:

magnetic trains works at the principle of repel on of the advantage is that they are fast and dont really need diesel

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