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4 years ago

List two possible reasons the engine oil could have a strong gasoline smell

1 answer:

8 0

Two reasons why the engine oil could have a strong gasoline smell is- why gas getting into the engine oil, is because your fuel mixture is too rich and If your fuel mixture is too rich, the combustion chamber won't ignite all of the fuel.

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While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),

MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , $y$ - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

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

$t$ - Time, measured in seconds.

If we know that $y_{o} = 2\,m$ , $y = 0\,m$ , $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then the time taken by the ball is:

$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t \approx 2.890\,s$ , then the final vertical velocity is:

$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball ( $x$ ) is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground ( $v$ ), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta_{o} = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , the angle of the total velocity of the ball just before hitting the ground is:

$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

3 0

3 years ago

Read 2 more answers

Antilaser eyewear should be worn when a laser level's output is greater than ____

creativ13 [48]

Answer: 5mW

Explanation:

8 0

3 years ago

If you had to pick a priority for future engineers, what would it be and why?

quester [9]

Answer:

Explanation:

Civil engineers have become experts in creating sustainable and environmentally friendly buildings and systems. Multiplied over many communities, the energy and emissions savings can make a real difference in the environment. Other life-improving functions can also make communities better places to live.Jul 19, 2017

7 0

3 years ago

Assume the average fuel flow rate at the peak torque speed (1500 rpm) is 15kg/hr for a sixcylinder four-stroke diesel engine und

sveticcg [70]

Answer:

Q = 8.845 DEGREE

Explanation:

given data:

combine Mass for 6 cylinder (M) =15 Kg/hr

mass of each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec

Engine speed (N)= 1500rpm

Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m

Discharge Coefficient (Cd) = 0.75

Pressure difference = 100 MPa

Density of fuel = 800 kg/m^3

velocity of fuel is $v = cd\sqrt{\frac{2*P}{p}}$

$v = 0.75 \sqrt{\frac{2\times 100\times 10^6}{800}} = 375 m/sec$

injected fuel volume (V) =Area of given Orifices × Fuel velocity × time of single injection × no of injection/sec

we know that p = m/ V

So $V = \frac{0.000694}{800} =8.68\times10^{-7} m3/sec$

putting these value in volume equation and solve for Discharge $8.68\times 10^{-7} = (\frac{(3.14}{4})\times 6\times( .0002\times .0002) \times 375 \times \frac{(Q}{360}) \times \frac{30}{750} \times \frac{(750}{60)}$

Q = 8.845 DEGREE

4 0

3 years ago

hree large plates are separated bythin layers of ethylene glycol and water. The top plate moves to the right at 2m/s. At what sp

MakcuM [25]

Answer: For the center plate to remain stationed in one position without rotating, the bottom plate has to move to the left at a speed of 2m/s, so as to cancel the force acting on it from the top.

The center plate will not move when the bottom plate is moving left in a speed of 2m/s to counter the speed of the top plate, because a body will continue to be at rest if all the forces acting towards the body are equal. The center plate will be at rest because we have directed equal force from the top and bottom of the plate.

3 0

3 years ago