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3 years ago

Considere que helicóptero para na altitude a que voava para largar uma caixa de 90 kg para o solo.

Physics

1 answer:

Firdavs [7]3 years ago

7 0

Answer:

El peso es de 882 newton.

Explicación:

El helicóptero se detiene a la altura en la que volaba para dejar caer una caja de 90 kg al suelo, el peso de la caja llega hasta los 882 Newton porque el peso es una fuerza dirigida hacia el centro de la tierra y depende de la gravedad. La fórmula del peso es multiplicar la masa por la gravedad, de modo que cuando se multiplican 90 kg de masa por 9,8 metros / seg2, obtenemos el peso del objeto que es 882 newton.

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What conditions must be satisfied for a body to be in equilibrium <br>

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3 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0

3 years ago

kramer

Answer:

60.2 J

Explanation:

Efficiency is the ratio of work out to work in.

e = Wout / Win

0.86 = Wout / 70 J

Wout = 60.2 J

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3 years ago

Read 2 more answers

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3 years ago

A deuteron, 21H, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a n

Naddik [55]

Answer:

1917723.40119 m/s

Explanation:

m = Mass of deuteron = $(1.637+1.675)\times 10^{-27}=3.312\times 10^{-27}$

k = Boltzmann constant = $1.381\times 10^{-23}\ J/K$

T = Temperature = $2.94\times 10^8\ K$

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8 0

3 years ago