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NISA [10]
3 years ago
6

Si se deja caer un carrito de una pista de coches sin friccion y su altura inicial es de 1.4 metros, cual es la velocidad maxima

que puede alcanzar el carrito?
Physics
1 answer:
Svet_ta [14]3 years ago
8 0

Answer:

5.241\ \text{m/s}

Explanation:

m = Masa del coche

g = Aceleración debida a la gravedad = 9.81\ \text{m/s}^2

h = Altura = 1.4\ \text{m}

v = Velocidad del automóvil en la parte inferior de la pista

Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.4}\\\Rightarrow v=5.241\ \text{m/s}

La velocidad máxima que puede alcanzar el coche es 5.241\ \text{m/s}.

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
Which of the follow represent a chemical changes
mixer [17]

Answer:

C) steel turning to rust in salt air

Explanation:

The missing options are:

A) ice melting to form liquid water

B) water boiling to form steam

C) steel turning to rust in salt air

D) sugar dissolving into hot coffee

In a chemical change the atoms of the reacting compounds are reordered forming new compounds. In a chemical change, new compounds appear, but in a physical change not.

Then, change of states like ice melting and water boiling are not chemical changes.

During steel rust, components of steel, like iron, are oxidized, that is, reacts with oxygen forming oxides.

The dissolution of sugar into hot coffee is a physical change in which sugar molecules get further apart in the coffee,  but they don't change.

5 0
3 years ago
In nuclear physics wht units are used to measure the radius of an atom ?
Alecsey [184]
Angstrom = 10^-10 m
for nucleus size are used fermi (femtometer  10^-15 m )
6 0
3 years ago
A- 1000 m/s2<br> Xi-0m<br> Xf-0.75m<br> Vf-?
sleet_krkn [62]

Answer:

The final velocity of the object is,  v_{f} = 27 m/s    

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, x_{i} = 0 m

The final displacement of the object,  x_{f} = 0.75 m

The initial velocity of the object will be, v_{i} = o m/s

The final velocity of the object, v_{f} = ?

The average velocity of the object,

                                    v = ( x_{f} - x_{i} )/ t

                                      = 0.75 / t

The acceleration is given by the relation

                                     a = v / t

                                   1000 m/s² = 0.75 / t²

                                            t² = 7.5 x 10⁻⁴

                                            t = 0.027 s

Using the I equation of motion,

                                  v_{f} = u + at

Substituting the values

                                   v_{f} = 0 + 1000 x 0.027

                                                           = 27 m/s

Hence, the final velocity of the object is,  v_{f} = 27 m/s          

8 0
3 years ago
An object is thrown with velocity v from the edge of a cliff above level ground. Neglect air resistance. In order for the object
musickatia [10]

Answer:

(C) greater than zero but less than 45° above the horizontal

Explanation:

The range of a projectile is given by R = v²sin2θ/g.

For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°

2θ = 90°

θ = 90°/2 = 45°

So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.

4 0
3 years ago
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