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wolverine [178]
3 years ago
10

Give two conditions in which an object becomes weightless​

Physics
2 answers:
stepan [7]3 years ago
7 0

Answer:

At the time of zero gravity there is no weight and at the time of where is there is no air and there is vaccum there is no weight

Explanation:

Evgesh-ka [11]3 years ago
4 0

Answer:

A body becomes weightless in a zero-gravity scenario and when a force is applied to a body that is equal and opposite to the force of gravity.

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8.34 m/s
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19. The work done by a magnetic field on a charged particle moving in it is:
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Answer:

The work done by a magnetic feild on a chraged particle moving in it is zero

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4 years ago
Dibuja la gráfica de calentamiento de un kilogramo de plomo que se encuentra inicialmente a 70ºC y pasa a una temperatura final
MariettaO [177]

Answer:

Q= m c_e ΔT and   Q = m L

Explanation:

For this graph of temperature vs energy (heating) we must use two relations

* for when there is no change of state

          Q= m c_e ΔT

* for using there is change of state

          Q = m L

the second expression is a consequence of the fact that all the energy supplied is used to change the state of the solid-liquid and liquid-gas system

the energy supplied is the sum of the energy in each interval

divide the system into intervals determined by the state change points

1) from T₀ = 70ºC to T_f = 327.4ºC, sample in solid-liquid state

           c_e = 128 J / kg ºC

           Q₁ = m c_e (T_f -To)

           Q₁=1  128 (327.4 -70)

           Q₁ = 3.29 10⁴ J

           Q = Q₁ = 3.29 10⁴ J

2) when is it changing from solid to liquid

            L = 2.45 10⁴ J / kg

            Q2 = 1 2.45 10⁴

            Q2 = 2.45 10⁴ J

            Q = Q₁ + Q₂

             Q = 5.74 10⁴ J

3) from to = 327.4ºC until T_f = 1725ºC, sample in liquid state

in the tables the specific heat of the solid and liquid state is the same

             Q3 = m c_e (T_f -To)

             Q3 = 1 128 (1725 -327.4)

             Q3 = 1.79 10⁵ J

              Q = Q₁ + Q₂ + Q₃

              Q = (3.29 +2.45 + 17.9) 10⁴ J

              Q = 23.64 10⁴ J

4) for when it is changing from the liquid state to the gaseous state

             L_v = 8.70 10⁵ J / kg

             Q₄ = m L_v

             Q₄ = 1 8.70 10⁵

             Q₄ = 8.70 10⁵ J

             Q = Q₁ + Q₂ + Q₃ + Q₄

              Q = (3.29 +5.74 + 17.9+ 87.0) 10⁴ J

               Q = 110.64 10⁴ J

5) from To = 1725ºC to T_f = 2000ºC, sample in gaseous state

             Q₅ = m c_e ΔT

             Q₅ = 1 128 (2000 -1725)

             Q₅ = 3.52 10⁴ J

             Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

              Q = 114.16 104 J

the following table shows the points to be plotted

         Energy (10⁴ J)  Temperature (ºC)

                  0                     70

                 3.29             327.4

                 5.74             327.4

               23.64           1725

               110.64          1725

                114.16         2000

In the attachment we can see a graph of Temperature versus energy supplied

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Why don't atoms get too close?
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Answer:

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A toy cannon uses a spring to project a 5.35-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force
Elenna [48]

Answer:

a) the velocity is v=1.385 m/s

b) the ball has its maximum speed at 4.68 cm away from its compressed position

c)  the maximum speed is 1.78 m/s

Explanation:

if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore

we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball

W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki

therefore

Ui + Ki = Uf + Kf + W fr  

where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.

since

U= 1/2 k x² , K= 1/2 m v²  , W fr = F*L

X= compression length , L= horizontal distance covered

therefore

Ui + Ki = Uf + Kf + W fr

1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L

a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0

1/2 k X² + 0 = 0 + 1/2 m v² + F*L

1/2 m v² = 1/2 k X² - F*L

v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s

b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.

1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]

1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d

taking the derivative

dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm

at x m = -0.4 cm the velocity is maximum

therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position

c) the maximum speed is

1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0

v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s

4 0
3 years ago
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