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3 years ago

gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure

Physics

1 answer:

Mamont248 [21]3 years ago

7 0

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be ' $\text{P}_2$ '.

As we know that Boyle's law formula states that;

$P_1 \times V_1 = P_2 \times V_2$

where, $P_1$ = original pressure of gas in the container = 310 mm hg

$P_2$ = required new pressure

$V_1$ = volume of gas in the container = 185 ml

$V_2$ = desired new volume of the gas = 74 ml

So, $P_2 = \frac{P_1 \times V_1}{V_2}$

$P_2 = \frac{310 \times 185}{74}$

= 775 mm hg

Hence, the required new pressure is 775 mm hg.

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A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl

USPshnik [31]

Given Information:

Wavelength = λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0

3 years ago

An object moves at 60 m/s in the +x direction. As it passes through the origin it gets a 4.5 m/s^2 acceleration in the -x direct

Brut [27]

Answer:

a) After 26.67 seconds it returns back to the origin

b) Velocity when it returns to the origin = 60 m/s in the -x direction

Explanation:

a) Let the starting position be origin and time be t.

After time t displacement, s = 0 m

Initial velocity, u = 60 m/s

Acceleration, a = -4.5 m/s²

We have equation of motion s = ut + 0.5 at²

Substituting

s = ut + 0.5 at²

0 = 60 x t + 0.5 x (-4.5) x t²

2.25t² - 60 t = 0

t² - 26.67 t = 0

t (t-26.67) = 0

t = 0s or t = 26.67 s

So after 26.67 seconds it returns back to the origin

b) We have equation of motion v = u + at

Initial velocity, u = 60 m/s

Acceleration, a = -4.5 m/s²

Time , t = 26.67

Substituting

v = 60 - 4.5 x 26.67 = -60 m/s

Velocity when it returns to the origin = 60 m/s in the -x direction

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4 years ago

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Maru [420]

Answer:

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7 0

3 years ago

(b) A ball is thrown from a point 1.50 m above the ground. The initial velocity is 19.5 m/s at

Nataly_w [17]

The answers are:

(i) 6.35m

(ii) 20.2 m/s

It seems like you already have the answer, but let me show you how to get it:

You have two givens:

Vi = 19.5m/s

Θ = 30°

dy = 1.50m (This is not the maximum height, just to be clear)

When working with these types of equations, you just need to know your kinematics equations. For projectiles launched at an angle, you will need to first break down the initial velocity (Vi) into its horizontal (x) and vertical (y) components.

*Now remember this, if you are solving for something in the horizontal movement always use only x-components. When solving for vertical movements, always use y-components.

Let's move on to breaking down the initial velocity into both y and x components.

Viy = SinΘVi = (Sin30°)(19.5m/s) = 9.75 m/s

Vix = CosΘVi = (Cos30°)(19.5m/s) = 16.89 m/s

Okay, so we have that down now. The next step is to decide which kinematics equation you will use. Because you have no time, you need to use the kinematic equation that is not time dependent.

(i) Maximum height above the ground

Remember that the object was thrown 1.50m above the ground. So we save that for later. First we need to solve for the maximum height above the horizontal, or the point where it was thrown.

The kinematics equation you will use is:

$Vf^{2} = Vi^{2}+2ad$

Where:

Vf = final velocity

Vi = inital velocity

a = 9.8m/s²

d = displacement

We will derive our displacement from this equation. And you will come up with this:

$d = \dfrac{Vf^{2}-Vi^{2}}{2a}$

Again, remember that we are looking for a vertical component or y-component because we are looking for HEIGHT. So we use this plugging in vertical values only.

Vf at maximum height is always 0m/s because at maximum height, objects stop. Also because gravity is a downwards force you will use -9.8m/s².

Vfy = 0 m/s a = -9.8m/s² Viy = 9.75m/s

$dy = \dfrac{Vfy^{2}-Viy^{2}}{2a}$

$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

$dy = \dfrac{-95.0625m^{2}/s^{2}}{-19.6m/s^{2}}$

$dy = 4.85m$

So from the point it was thrown, it reached a height of 4.85m. Now we add that to the height it was thrown to get the MAXIMUM HEIGHT ABOVE THE GROUND.

4.85m + 1.50m = 6.35m

(ii) Speed before it strikes the ground. (Vf=resultant velocity)

Okay, so here we need to consider a couple of things. To get the VF we need to first figure out the final velocities of both the x and y components. We are combining them to get the resultant velocity.

Vfx = horizontal velocity = Initial horizontal velocity (Vix). This is because gravity is not acting upon the horizontal movement so it remains constant.

Vfy = ?

VF =?

We need to solve this, again, using the same formula, but this time, you need to consider we are moving downwards now. So this time, instead of Vfy being 0 m/s, Viy is now 0 m/s. This is because it started moving from rest.

$Vfy^{2} = Viy^{2}+2ad$