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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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Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is
victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0
3 years ago
A 25 kg child runs at a speed of 5.0 m/s and jumps onto a stationary shopping cart and holds on for dear life. The cart has mass
makkiz [27]

Answer:

3.38 m/s

Explanation:

Mass of child = m₁ = 25

Initial speed of child = u₁ = 5 m/s

Initial speed of cart = u₂ = 0 m/s

Mass of cart = m₂ = 12 kg

Velocity of cart with child on top = v

This is a case of perfectly inelastic collision

m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\frac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\frac{25\times 5+12\times 0}{25+12}\\\Rightarrow v=\frac{125}{37}\\\Rightarrow v=3.38\ m/s

Velocity of cart with child on top is 3.38 m/s

7 0
3 years ago
Can you hellp me please
Agata [3.3K]

Answer:

The answer is D.

Explanation: The northern hemisphere is more cooler, because it doesn't have much sunlight to warm it up.

6 0
3 years ago
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A circuit contains a single 220 pF capacitor hooked across a battery. It is desired to store three times as much energy in a com
Anettt [7]

Answer

The capacitor should be connected in parallel as parallel connection gives the arithmetic sum of capacitance which will give a corresponding sum of energy while capacitors in series gives the sum of the reciprocal if the individual capacitance

7 0
3 years ago
Find the y-component of this
Alborosie

Answer:

-0.0789 m

Explanation:

Recall that the y-component comes associated with the sin(18.4) through the following trigonometric relationship:

y = 0.250 sin(-18.4) ≈ -0.0789 m

Notice it is negative since it is below the x-axis.

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