PHYSICS COLLEGE

Trevor and Nick are taxi drivers. Trevor drives a taxi using diesel oil while Nick drives a taxi using LPG. Whose taxi will cause higher levels of air pollution and why?

Answers

Answer 1

Answer:

Answer:

Trevor's taxi would cause higher levels of air pollution

Explanation:

Trevor's taxi use diesel oil.

Diesel is less cleaner than LPG.

Compared to automotive pollution from petrol and diesel, pollutants from LPG-driven cars include lower amounts of petroleum hydrocarbons, nitrogen oxides , sulphur oxides, ozone contamination and particulate matter.

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A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross- sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum flux correction factor to be 1.03 at both the inlet and outlet.

Answers

The anchoring force needed to hold the elbow in place is 839.5 N.

How to find anchoring force?

To determine the anchoring force needed to hold the elbow in place, use the following steps:

Apply the conservation of momentum equation to the elbow:

∑F = ˙m(V₂ - V₁)

where:

˙m = mass flow rate

V₁ and V₂ = velocities at the inlet and outlet of the elbow, respectively

F = anchoring force

The mass flow rate is given by:

˙m = ρAV

where:

ρ = density of water (1000 kg/m³)

A = cross-sectional area of the elbow

V = velocity

The velocities at the inlet and outlet of the elbow can be calculated using the following equations:

V₁ = A₁V₁

V₂ = A₂V₂

where:

A₁ and A₂ = cross-sectional areas at the inlet and outlet of the elbow, respectively

Calculate the momentum flux correction factor at the inlet and outlet of the elbow:

β₁ = 1.03

β₂ = 1.03

Substitute all of the above equations into the conservation of momentum equation:

F = ˙m(V₂ - V₁)

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

Calculate the velocity at the inlet of the elbow:

V₁ = A₂V₂/A₁

V₁ = (25 cm²/150 cm²)(V₂)

V₁ = 1/6 V₂

Substitute the above equation into the conservation of momentum equation:

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

F = ρA₁[(1/6 V₂)²](1.03) - ρA₂V₂²(1.03)

F = ρV₂²(1.03)(1/36 A₁ - A₂)

Calculate the anchoring force:

F = (1000 kg/m³)(V₂²)(1.03)(1/36 × 150 cm² - 25 cm²)

F = 839.5 N

Therefore, the anchoring force needed to hold the elbow in place is 839.5 N.

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The anchoring force needed to hold the elbow in place is 932 N.

The anchoring force needed to hold the elbow in place is the sum of the following forces:

The force due to the change in momentum of the water as it flows through the elbow.

The force due to the weight of the elbow and the water in it.

The force due to the buoyancy of the water in the elbow.

The force due to the change in momentum of the water can be calculated using the momentum equation:

F = mΔv

where:

F is the force

m is the mass of the fluid

Δv is the change in velocity of the fluid

In this case, the mass of the fluid is the mass of the water that flows through the elbow per second. This can be calculated using the mass flow rate equation:

m = ρAv

where:

ρ is the density of the fluid

A is the cross-sectional area of the pipe

v is the velocity of the fluid

The velocity of the fluid at the inlet can be calculated using the Bernoulli equation:

P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2

where:

P1 is the pressure at the inlet

v1 is the velocity at the inlet

P2 is the pressure at the outlet

v2 is the velocity at the outlet

In this case, the pressure at the outlet is atmospheric pressure. The velocity at the outlet can be calculated using the continuity equation:

A1v1 = A2v2

where:

A1 is the cross-sectional area at the inlet

A2 is the cross-sectional area at the outlet

The force due to the weight of the elbow and the water in it is simply the weight of the elbow and the water in it. The weight can be calculated using the following equation:

W = mg

where:

W is the weight

m is the mass

g is the acceleration due to gravity

The force due to the buoyancy of the water in the elbow is equal to the weight of the water displaced by the elbow. The weight of the water displaced by the elbow can be calculated using the following equation:

B = ρVg

where:

B is the buoyancy

ρ is the density of the fluid

V is the volume of the fluid displaced

g is the acceleration due to gravity

The volume of the fluid displaced by the elbow is equal to the volume of the elbow.

Now that we have all of the forces, we can calculate the anchoring force needed to hold the elbow in place. The anchoring force is equal to the sum of the forces in the negative x-direction. The negative x-direction is the direction in which the water is flowing.

F_anchor = F_momentum + F_weight - F_buoyancy

where:

F_anchor is the anchoring force

F_momentum is the force due to the change in momentum of the water

F_weight is the force due to the weight of the elbow and the water in it

F_buoyancy is the force due to the buoyancy of the water in the elbow

Plugging in the values for each force, we get:

F_anchor = 1030 N - 490 N + 392 N = 932 N

Therefore, the anchoring force needed to hold the elbow in place is 932 N.

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A sphere has a charge of −84.0 nC and a radius of 5.00 cm. What is the magnitude of its electric field 3.90 cm from its surface?

Answers

Answer:

$E = -9.5* 10^4~N/C$

Explanation:

Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.

In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.

Gauss' Law states that

$\int {\vec{E}d\vec{a}} = (Q_(enc))/(\epsilon_0)$

Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.

So,

$E4\pi r^2 = (-84* 10^(-9))/(8.8* 10^(-12))\nE4\pi (8.9 * 10^(-2))^2 = (-84* 10^(-9))/(8.8* 10^(-12))\n\nE = -9.5* 10^4~N/C$

A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.

Answers

Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

$\Sigma F = F - m\cdot g = m\cdot a$ (1)

Where:

$F$ - Buoyant force, measured in newtons.

$m$ - Mass of the plastic ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration, measured in meters per square second.

If we know that $F = 5\,N$ , $m = 0.408\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then the net acceleration of the plastic ball is:

$a = (F)/(m) - g$

$a= 2.448\,(m)/(s^(2))$

The acceleration is 2.448 meters per square second and is vertically upward.

when a ball is dropped off a cliff in free fall, it has an acceleration of 9.8 m/s^2. what is its acceleration as it gets closer to the ground

Answers

acceleration due to gravity is contract for the purposes of this question, so the acceleration would remain at 9.8 m/s^2

If the ball, the cliff, and the ground are all on the Earth, and everything is bathed in an ocean of air, then the ball's acceleration will decrease as it falls, because of the friction of air resistance. If it has far enough to fall, it's possible that its acceleration may even become zero, and the ball settle on a constant speed (called "terminal velocity") before it hits the ground.

But until we get to College-level Physics and Engineering, we ALWAYS ignore that stuff, and assume NO AIR RESISTANCE. The ball is in FREE FALL, and the ONLY force acting on it is the force of gravity. We also assume that the distance of the fall is small enough so that the value of gravity is constant over the entire fall.

Under those assumptions, there's nothing present to change the acceleration of the falling ball. It's 9.81 m/s² when it rolls off the edge of the cliff, and it's 9.81 m/s² when it hits the ground.

Early in the morning, when the temperature is 5.5 °C, gasoline is pumped into a car’s 53-L steel gas tank until it is filled to the top. Later in the day the temperature rises to 27 °C. Since the volume of gasoline increases more for a given temperature increase than the volume of the steel tank, gasoline will spill out of the tank. How much gasoline spills out in this case?

Answers

Answer:

Volume of gasoline spills out is 0.943 L.

Explanation:

Volumetric expansion of both gasoline and steel tank is :

$\beta_(gas)=9.5 *10^(-4)/K\n\beta_(steel \ gas)=3.6 * 10^(-5)/K.$ { source Internet}

We know expansion due to temperature change is :

$\Delta V=\beta*\Delta T* V$

For gasoline:

$\Delta V_g=0.98 \ L.\n$

Similarly for Steel tank:

$\Delta V_(steel \ gas)=0.037\ L$ .

Now, volume of gasoline spills out is equal to difference between expansion in volume.

$\Delta V_(gas)-\Delta V_(Steel \ gas)=0.98-0.037\ L=0.943\ L.$

Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horizontally with a constant speed of 3.80 m/s relative to observer 2 standing on the sidewalk.a) What is the speed of the ball 1.00 s after it is released, as measured by observer 2?

b) What is the direction of travel of the ball 1.00 s after it is released, as measured relative to the horizontal by observer 2?

Answers

a) 10.5 m/s

While for observer 1, in motion with the car, the ball falls down straight vertically, according to observer 2, which is at rest, the ball is also moving with a horizontal speed of:

$v_x = 3.80 m/s$

As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

$v_y = u_y + gt$

where

$u_y =0$ is the initial vertical speed

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

t is the time

Therefore, after t = 1.00 s, the vertical speed is

$v_y = 0 + (9.8)(1.00)=9.8 m/s$

And so the speed of the ball, as observed by observer 2 at rest, is given by the resultant of the horizontal and vertical speed:

$v=√(v_x^2 +v_y^2)=√((3.8)^2+(9.8)^2)=10.5 m/s$

b) $\theta = -68.8^(\circ)$

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

$\theta = tan^(-1) ((v_y)/(v_x))=tan^(-1) ((-9.8)/(3.8))=-68.8^(\circ)$

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so $v_x$ has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so $v_y$ is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

$\theta = -68.8^(\circ)$

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