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netineya [11]
4 years ago
12

The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on convention

al gasoline vehicles.
Engineering
1 answer:
gulaghasi [49]4 years ago
7 0

Answer:

Batteries are safe when handled properly.

Explanation:

Just like the battery in your phone, the battery in some variant of an electric car is just as safe. If you puncture/smash just about any common kind of charged battery, it will combust. As long as you don't plan on doing anything extreme with the battery (or messing with high voltage) you should be fine.

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A particle of given mass m 1kg, and charge q1 Couloumb, is placed in a magnetic field B 1k Tesla and thus is subjected to the ef
GalinKa [24]

Answer:

a) y = -0.05*x^2

b) Range shorter, Radius of circular/spiral motion shorter

Explanation:

Given:

- The charge q = 1 C

- Magnetic Field = 1 k Tesla

- Mass of the charge m = 1 kg

- Acceleration due to Lorentz Force a = q ( v x B ) / m

- Acceleration due to Drag Force a = - c*v

- The velocity of the charge @ x = y = 0, v_i = 10 i

Find:

- Equation of path of the charge in the given magnetic field

- Describe the path with addition of Drag Force.

Solution:

- First we will determine the magnitude and direction of the acceleration due to Lorentz Force as follows:

                                  a =  q ( v x B ) / m

- Input values:

                                  a =  1 ( 10 i x 1 k ) / 1 = -10 j

- We see that the acceleration acts in the negative y direction.

- To determine the path of the charge, we will use kinematic equation of motion for the particle as follows:

                                   x =  v_i*t

                                   y = 0.5*a*t^2

- Determine the parametric equations for displacements in x and y directions:

                                    x = 10*t

                                    y = -0.5*10*t^2

Combine the parametric equations to determine the equation of path followed by the charge:

                                    t = x / 10

                                    y = -0.5*x^2 / 100

                                    y = -0.05*x^2

- For the second case we have the acceleration due to drag force relation:

                                    a = - 0.1*v_i

                                    a = -0.1*10 i

                                    a = - i

- There is a deceleration component of drag force acting in x direction, The parametric equation would be as follows:

                                    x = 10*t - 0.5*1*t^2

                                    x = 10*t - 0.5*t^2

- From this we can see that the x coordinate increases at a decreasing rate. Hence, the range of the projectile like motion will decrease. The total amount of distance traveled by the charge in x direction will decrease.

-Hence, after sufficient amount of time the charge moves in a circular motion. The radius of the circular motion will be shorter as compared to when we neglect the acceleration due to drag force.

5 0
3 years ago
26 points!!
Alexandra [31]
Answer:
Tesla switched to smaller front motors to permanent magnets on the model S and X after seeing the mileage results of model 3 permanent magnet motors.

The model 3 got more mileage but they make their money by keeping people coming back for a new car after their mileage runs out.

So they switched to a less quality standard model.
8 0
3 years ago
Hi all any one help me?? ​
yanalaym [24]

Answer:

Explanation:

sorry i dont know

5 0
3 years ago
Read 2 more answers
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar. 

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0
4 years ago
Can someone please answer this, it will be very appreciative
Lisa [10]

Answer:

sorry the image isn't clear

Explanation:

7 0
2 years ago
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