Please help due today

A bicycle pump contains 20 cm3 of air at a pressure of 100 kPa. The air is then pumped in a tyre of volume 100 cm3. Calculate th

Natasha2012 [34]

Answer:

The pressure of the air in the tyre is 20 kPa

Explanation:

The parameters for the bicycle pump and tyre are;

The volume of air contained in the bicycle pump, V₁ = 20 cm³

The pressure of the air contained in the bicycle pump, P₁ = 100 kPa

The volume (available) of the tyre, where the air is pumped, V₂ = 100 cm³

Let P₂ represent the pressure in the tyre after the air is pumped

By Boyle's law, we have that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure;

Mathematically, Boyle's law gives the following equation;

P₁ × V₁ = P₂ × V₂

∴ P₂ = (P₁ × V₁)/V₂

Substituting the known values gives;

P₂ = (100 kPa × 20 cm³)/(100 cm³)

∴ P₂ = 100 kPa × 1/5 = 20 kPa

P₂ = 20 kPa

The pressure of the air in the tyre = P₂ = 20 kPa.

7 0

3 years ago

Suppose that the collector is held at a small negative voltage with respect to the grid. Will the accelerated electrons reach th

Leona [35]

Answer:

B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.

Explanation:

In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential

so according to the given situation, the option b is correct

And, the rest of the options are wrong

3 0

3 years ago

6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri

mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a = - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

Read 2 more answers

a car collides with a wall. compare the forces exerted by the car on the wall and the wall on the car

NeTakaya

Mass multiplied by acceleration produces force.

The acceleration is (v - 0)/t in this situation, where t seems to be the time it takes automobile A to come to a stop. According to Newton's third law of motion, the automobile produces this turning force of the wall, however the wall, which really is static and indestructible, forces an equal force back on the car.

According to Newton's third law, each action has an equal and opposite response. On this basis, you may deduce that a car driving into a wall would exert force on the wall. However, since the wall did not move, the automobile receives an equivalent force, causing it to collapse.

<h3>Learn more:</h3>

brainly.com/question/13952508?referrer=searchResults

5 0

2 years ago

Read 2 more answers