In a rigid container, when the speed of the gas molecules decreases, the pressure of the gas also decreases.

Renewable resources need to be conserved because

creativ13 [48]

Answer:

(A) We are using them faster than they are replenished by nature

4 0

3 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

$v_x = 18.5 m/s$

we need to find the vertical velocity. This can be done by using the equation

$v_y = u_y -gt$

where

$u_y =0$ is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

$v_y = -(9.8)(3.03)=-29.7 m/s$

So the magnitude of the velocity at the impact (so, the speed at the impact) is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s$

The angle instead can be found as:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}$

so, 58.1 degrees below the horizontal.

4 0

3 years ago

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

$\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL$

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0

3 years ago

Kepler's third law is founded on a mathematical formula that is based on the inverse relationship between a planet's orbital vel

Cloud [144]

Kepler's third law is founded on a mathematical formula that is based on
the inverse relationship between a planet's orbital velocity and its distance
from the sun.

4 0

4 years ago

Read 2 more answers

Sound waves travel through air at a speed of 330 m/s. A whistle blast at a frequency of about 1.0 kHz lasts for 2.0 s. (a) Over

klasskru [66]

Answer:

a) x = 660 m , b) λ = 0.330 m , c) precision is 0.1 cm , d) Δf= n Δt/ t²

Explanation:

a) the speed of sound is constant, therefore we can use the relation of motion to inform the distance that the sound extends is

v = x / t

x = v t

x = 330 2

x = 660 m

b) the speed of sound is

v = λ f

λ = v / f

λ = 330/1000

λ = 0.330 m

c) a measuring tape must be used to measure the wavelength, the precision is 0.1 cm

.d) frequency measurement is more delicate, a stopwatch should be used to measure a certain number of oscillations, and hence calculate the frequency

.f = n / t

Therefore, if we assume that there is no error in the number of oscillations, the pressure is given by the appreciation of the stopwatch, which is maximum 0.01 s

Δf / f = Δt / t

Δf = Δt /t f

Δf= n Δt/ t²

3 0

4 years ago