Question 9

What falls faster? Light or heavy objects?

matrenka [14]

Heaver objects because they have greater mass

5 0

4 years ago

What might be done to prevent acid rain damage to objects made from metal and carbonate, such as limestone? Identify a metal obj

serg [7]

Answer: Bronze is a metal that can be damaged in acid rain. It is harsh to metals by almost burning them and may create rust. Everbrite is a clear, protective coating that will seal metal and bronze.

Explanation:

I know this is correct i have done all the research

7 0

3 years ago

A 1-m3 tank containing air @ 25 oC & 500 kPa is connected to another tank containing 5 kg of air at 35 oC & 200 kPa thro

VladimirAG [237]

Answer:

Volume of Tank 2, V' = $2.17 m^{3}$

Equilibrium Pressure, $P_{eq} = 278.82 kPa$

Given:

Volume of Tank 1, V = 1 $m^{3}$

Temperature of Tank 1, T = $25^{\circ}C$ = 298 K

Pressure of Tank 1, P = 500 kPa

Mass of air in Tank 2, m = 5 kg

Temperature of tank 2, T' = $35^{\circ}C$ = 303 K

Pressure of Tank 2, P' = 200 kPa

Equilibrium temperature, $20^{\circ}C$ = 293 K

Solution:

For Tank 1, mass of air in tank can be calculated by:

PV = m'RT

$m' = \frac{PV}{RT}$

$m' = \frac{500\times 1}{0.287\times 298} = 5.85 kg$

Also, from the eqn:

PV' = mRT

V' = volume of Tank 2

Thus

V' = $\frac{mRT}{P}$

V' = $\frac{5\times 0.287\times 303}{200} = 2.17 m^{3}$

Now,

Total Volume, V'' = V + V' = 1 + 2.17 = 3.17 $m^{3}$

Total air mass, m'' = m + m' = 5 + 5.85 = 10.85 kg

Final equilibrium pressure, P'' is given by:

$P_{eq}V'' = m''RT_{eq}$

$P_{eq} = \frac{m''RT_{eq}}{V''}$

$P_{eq} = \frac{10.85\times 0.87\times 293}{3.17} = 287.82 kPa$

$P_{eq} = 287.82 kPa$

3 0

3 years ago

Desde el balcón de un edificio se deja caer una manzana y llega a la planta baja en 5 s. ¿Desde qué piso se dejó caer, si cada p

miv72 [106K]

Answer: 42

Explanation:

I will answer this in English.

We know that the apple needs 5 seconds to reach the ground.

Each floor of the building has a height of 2.88m.

Now, when we drop something, the only force acting on the object is the gravitational one, so the acceleration of the apple is:

a = -g

for the velocity, we integrate the acceleration over time, and as the apple is dropped, we do not have any initial velocity, so we do not have a constant in the integration:

v = -g*t

for the position we integrate again, now we have an initial height H, so the position is:

p = (-g/2)*t^2 + H

now the apple hits the ground when p = 0, so we can solve this equation to find H.

i will use g = 9.8m/s^2

0 = (-4.9m/s^2)*(5s)^2 + H

H = 122.5 m

now knowing H, we can divide it by the height of a floor in the building and get the number of the floor.

N = 122.5m/2.88m = 42.5

this means that the apple was dropped in the floor 42 (the 0.5 means that the apple was not right where the floor 42 starts, it was dropped around the middle of the floor 42)

5 0

3 years ago

The 20 g bullet is traveling at 400 m/s when it becomes embedded in the 2 -kg stationary block. Determine the distance the block

Marina CMI [18]

Answer:

The block+bullet system moves 4 m before being stopped by the frictional force.

Explanation:

Using the law of conservation of llinear momentum and the work energy theorem, we can obtain this.

According to Newton's second law of motion

Momentum before collision = Momentum after collision

Momentum before collision = (0.02×400) + 0 (stationary block)

Momentum before collision = 8 kgm/s

Momentum after collision = (2+0.02)v

8 = 2.02v

v = 3.96 m/s.

According to the work-energy theorem,

The kinetic energy of the block+bullet system = work done by Friction to stop the motion of the block+bullet system

Kinetic energy = (1/2)(2.02)(3.96²) = 15.84 J

Work done by the frictional force = F × (distance moved by the force)

F = μmg = 0.2(2.02)(9.8) = 3.96 N

3.96d = 15.84

d = (15.84/3.96) = 4 m

5 0

3 years ago