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2 years ago

Cual es la energía cinética de un objeto cuya masa es de 2kg y se mueve a una velocidad de 5 m/s2 * a)10 J b)25 J c)50J d)98J

1 answer:

7 0

Let the mass be m
v = 5m/s
K.E = 25 J
K.E = 1/2 mv^2
25 = 1/2 m (5)^2
25 x 2 = 25m
m = 2 kg
Now if the velocity becomes 3 times then the velocity is 3 x 5 = 15m/S
Now new K.E = 1/2 x 2 x 15^2
= 225 J

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A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj

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2 years ago

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3 years ago

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The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

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To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

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$|\Delta E_{ij}| = |E_i-E_j|$

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In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

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Therefore the states of energy would be

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3 years ago

On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze

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Answer:

0.78333 m/s in the opposite direction

1.566 m/s in the same direction

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$m_1$ = Mass of penny = 0.0025 kg

$m_2$ = Mass of nickel = 0.005 kg

$u_1$ = Initial Velocity of penny = 2.35 m/s

$u_2$ = Initial Velocity of nickel = 0 m/s

$v_1$ = Final Velocity of penny

$v_2$ = Final Velocity of nickel

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s$

The final velocity of the penny is 0.78333 m/s in the opposite direction

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s$

The final velocity of the nickel is 1.566 m/s in the same direction

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2 years ago