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3 years ago

What is the potential energy possessed by a rock of mass 4kg projected to a height of 2m (g=10m/s2)

Physics

1 answer:

Mumz [18]3 years ago

5 0

Potential Energy = Mass * Gravitational Constant * Height

What We Don't Know: Potential Energy

What We Know:

Mass

Gravitational Constant (9.8 m/s)

Height

Ep = mgh

Ep = 4 * 9.8 * 2

Ep = 78.4 Joules

The Potential Energy of the Rock is 78.4 Joules, based on the given information! Hope this helps you!

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A wall has inner and outer surface temperatures of 25◦C and 8◦C, respectively. The interior and exterior air temperatures are 35

Angelina_Jolie [31]

Answer:

a) $\frac{\dot Q}{A} =60\ W.m^{-2}$

b) $\frac{\dot Q}{A} =110\ W.m^{-2}$

c) The wall may not be under steady because the two surfaces of the wall are exposed to the air at different temperatures and they have different convective coefficient.

Explanation:

Given:

temperature of the inner surface of the wall, $T_i=25^{\circ}C$

temperature of the outer surface of the wall, $T_o=8^{\circ}C$

temperature of the air outside, $T_{ao}=-3^{\circ}C$

temperature of the air inside, $T_{ai}=35^{\circ}C$

coefficient of heat convection on outside, $h_o=10\ W.m^{-2}.K^{-1}$

coefficient of heat convection on inside, $h_i=6\ W.m^{-2}.K^{-1}$

The heat flux between the interior air and the wall:

The convective heat transfer rate is given as,

$Q=h_i.A.\Delta T$

$\Rightarrow \frac{\dot Q}{A} =h_i\times (T_{ai}-T_i)$

$\frac{\dot Q}{A} =6\times (35-25)$

$\frac{\dot Q}{A} =60\ W.m^{-2}$

The heat flux between the exterior air and the wall:

$\Rightarrow \frac{\dot Q}{A} =h_o\times (T_{ao}-T_i)$

$\frac{\dot Q}{A}=10\times (8-(-3))$

$\frac{\dot Q}{A} =110\ W.m^{-2}$

The wall may not be under steady because the two surfaces of the wall are exposed to the air at different temperatures and they have different convective coefficient.

4 0

4 years ago

Pooping in my room and my room is upstairs and upstairs bathroom upstairs

diamong [38]

Answer:

huh? do you need help on math?

Explanation:

what do you mean?

7 0

2 years ago

Cannonball a is launched across level ground with speed v₀ at an angle of 45. Cannonball b is launched from the same location wi

postnew [5]

Answer:

Cannonball b spends more time in the air than cannonball a.

Explanation:

Starting with the definition of acceleration, we have that:

$a=\frac{\Delta v}{\Delta t}\\\\\Delta t= \frac{\Delta v}{a}$

Since both cannonballs will stop in their maximum height, their final velocity is zero. And since the acceleration in the y-axis is g, we have:

$\Delta t= -\frac{v_{oy}}{g}$

Now, this time interval is from the moment the cannonballs are launched to the moment of their maximum height, exactly the half of their time in the air. So their flying time t_f is (the minus sign is ignored since we are interested in the magnitudes only):

$t_f=2\frac{v_{oy}}{g}$