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2 years ago

Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface

Physics

2 answers:

ss7ja [257]2 years ago

7 0

Answer:

The total energy is 1080 J

Explanation:

KE = 1/2 . m . v^2 = 0.5 . 2 . 10^2 = 100J

Height energy ( relative to the surface) = m . g . h = 2 . 9.8 . 50 = 980 J

See: https://answers.yahoo.com/question/index?qid=20110215125659AAT5Q0M

mina [271]2 years ago

5 0

We have:

Total Energy: KE + GPE
KE (Kinetic Energy) = $\frac{1}{2} m*v^2$
GPE (Gravitational Potential Energy) = $m*g*h$

Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²

Formula:

Total Energy: KE + GPE
$TE = \frac{1}{2} m*v^2 + m*g*h$

Solving:
$TE = \frac{1}{2} m*v^2 + m*g*h$
$TE = \frac{1}{2} *2.0*10^2 + 2.0*10*50$
$TE = \frac{2.0*100}{2} + 1000$
$TE = \frac{200}{2} + 1000$
$TE = 100 + 1000$
$\boxed{\boxed{TE = 1100\:Joule}}\end{array}}\qquad\quad\checkmark$

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Answer:

A.The vertical velocity is constantly increasing as the ball falls.

B.The horizontal velocity does not noticeably change as the ball falls.

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Explanation:

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3 years ago

What is the full meaning of (i.p.s.m.n)

Anna11 [10]

Answer:

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3 years ago

A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti

Papessa [141]

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

$\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 - \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J$

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2 years ago

Read 2 more answers

Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a

nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

p₀o = m₁ v₁ + m₂ v₂

pf = 0

m₁ v₁ + m₂ v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂= - (-6.2) / 4.7

m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

Xcm = 1/M (m₁ x₁ + m₂ x₂)

We derive from time

Vcm = 1/M (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

0 = 1/M (m₁ v₁ + m₂v₂)

m₁ v₁ + m₂v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂ = 1.3

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jonny [76]

Answer:

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