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2 years ago

Depending on the model, all-electric vehicles can go about _______ miles on a single charge.

Engineering

1 answer:

Lera25 [3.4K]2 years ago

6 0

Answer:

Explanation:

Most electric vehicles can go at least 100.....few, if any, can go 400 or more on a single charge

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As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de

katen-ka-za [31]

yes it will

Explanation:

5 0

3 years ago

A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i

dlinn [17]

Answer:

$h_{max} = 51.8 cm$

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

$\theta$ = water surface angle with horizontal surface

$a_x =$ accelration in x direction

$a_z =$ accelration in z direction

slope in xz plane is

$tan\theta = \frac{a_x}{g +a_z}$

$tan\theta = \frac{4}{9.81+0}$

$tan\theta =0.4077$

the maximum height of water surface at mid of inclination is

$\Delta h = \frac{d}{2} tan\theta$

$=\frac{0.4}{2}0.4077$

$\Delta h 0.082 cm$

the maximu height of wwater to avoid spilling is

$h_{max} = h_{tank} -\Delta h$

= 60 - 8.2

$h_{max} = 51.8 cm$

the height requird if no spill water is $h_{max} = 51.8 cm$

3 0

3 years ago

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

What is the rated power output in ( kw) of a 8 pole motor designed to an IEC 180L motor frame ?<br>

ra1l [238]

Answer:

P=11 kW

Explanation:

Given that

Number of poles= 8

I.E.C. 180L motor frame

From data book , for 8 poles motor at 50 Hz

Speed = 730 rpm

Power factor = 0.75

Efficiency at 100 % load= 89.3 %

Efficiency at 50 % load= 89.1 %

Output power = 11 kW

Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.

P=11 kW

4 0

3 years ago

The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i

koban [17]

Answer: (a) 9.00 Mega Newtons or 9.00 * 10^6 N

(b) 17.1 m

Explanation: The length of wall under the surface can be given by

$b=25m/sin(60)\\=28.867$

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

$F(resultant) = Pavg ( A) = (Patm + \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N$

Noting from the Bernoulli equation that

$Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\$

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

$Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m$

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0

3 years ago