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3 years ago

5

Un objeto con una carga de 7μC se coloca en un campo eléctrico uniforme de 200N/C con dirección vertical. ¿Cuál es el valor de l

a masa del objeto si se queda "flotando" en el campo?

1 answer:

Hitman42 [59]3 years ago

5 0

Answer:

m = 1,429 10⁻⁴ kg

Explanation:

For the object to remain floating, the translational equilibrium condition must be met, we set a reference system where the vertical upward direction is positive

F_e -W = 0

F_e = W

q E = m g

we clear

m = $\frac{qE}{g}$

let's calculate

m = $\frac{7 \ 10^{-6} \ 200}{ 7 }$

m = 1,429 10⁻⁴ kg

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How much work is required to lift a 2 kilogram mass to a height of 10 meters

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Given mass = 2kg, height = 10m,g = 9.8.

We know that Work done W = FD

= > W = (mg)(D)

= > W = (2 * 9.8)(10)

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4 years ago

Read 2 more answers

A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal

Natalija [7]

Answer : The specific heat of unknown sample is, $8748.78J/kg^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-[q_2+q_3]$

$m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]$

where,

$c_1$ = specific heat of unknown sample = ?

$c_2$ = specific heat of water = $4186J/kg^oC$

$c_3$ = specific heat of copper = $390J/kg^oC$

$m_1$ = mass of unknown sample = 72.0 g = 0.072 kg

$m_2$ = mass of water = 203 g = 0.203 kg

$m_2$ = mass of copper = 187 g = 0.187 kg

$T_f$ = final temperature of calorimeter = $39.4^oC$

$T_1$ = initial temperature of unknown sample = $80.0^oC$

$T_2$ = initial temperature of water and copper = $11.0^oC$

Now put all the given values in the above formula, we get

$0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]$

$c_1=8748.78J/kg^oC$

Therefore, the specific heat of unknown sample is, $8748.78J/kg^oC$

7 0

3 years ago

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t

castortr0y [4]

Answer:

4.86 seconds

Explanation:

Velocity of projection, u = 14 m/s

angle of projection, θ = 20°

Formula for the time of flight

$T=\frac{2uSin\theta }{g}$

For earth

Te = (2 x 14 x Sin 20) / 9.8

Te = 0.98 s

For moon

g' = g/6 = 1.64 m/s^2

Tm = ( 2 x 14 x Sin 20) / 1.64

Tm = 5.84 seconds

Tm - Te = 5.84 - 0.98 = 4.86 s

So, it takes 4.86 s more time of flight on moon than the earth.

5 0

3 years ago