All categories

3 years ago

Which items do NOT describe possible solutions to air pollution?

Physics

1 answer:

bonufazy [111]3 years ago

6 0

Answer:

2, 6

Explanation:

2 because if you cut down more trees you will have less items to help collect co2

6 because if you have more manufracturing more gasses will be release and moe carbon dioxcide in the air so it will slowly kill the ozone layer.

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List three ways in which the study of science has made modern life different from that of 100 years ago

stealth61 [152]

Life Expectancy Was Shorter
Trains were faster to ride
The population explosion was not as on date

8 0

2 years ago

Biker A is cruising at a speed of 10.0 m/s when she passes biker B who is at rest at the origin of the coordinate system. Biker

Burka [1]

Answer: attached

Explanation:

3 0

3 years ago

A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia

Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0

3 years ago

What do we mean by principle of conservation?

leonid [27]

Answer:

Conservation principles tell us that some quantity, quality, or aspect remains constant through change.

Explanation:

7 0

3 years ago

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The

ikadub [295]

Answer:

A. $W=600\ J$

B. $Q=2112\ J$

C. $\Delta U=1512\ J$

D. $W=0\ J$

Explanation:

Given:

no. of moles of oxygen in the cylinder, $n=0.2$

initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

According to the question the final volume becomes twice of the initial volume.

Using ideal gas law:

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

$V_i=0.003\ m^3$

Work done by the gas during the initial isobaric expansion:

$W=P.dV$

$W=P_i\times (V_f-V_i)$

$W=2\times 10^5\times (0.006-0.003)$

$W=600\ J$

we have the specific heat capacity of oxygen at constant pressure as:

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

$\frac{0.003}{360} =\frac{0.006}{T_f}$

$T_f=720\ K$

Now change in internal energy:

$\Delta U=n.c_p.(T_f-T_i)$

$\Delta U=0.2\times 21\times (720-360)$

$\Delta U=1512\ J$

Now heat added to the system:

$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

$W=0\ J$

7 0

3 years ago