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riadik2000 [5.3K]
4 years ago
11

Technician A says to use​ lineman's gloves when servicing the high voltage system on hybrid electric vehicles. Technician B says

that hybrid vehicles have a plug to disable the high voltage system before service. Which technician is​ correct?
Physics
1 answer:
djyliett [7]4 years ago
5 0

Answer:

Both technicians are correct

Explanation:

Insulated rubber gloves or lineman’s gloves are a major requirement to ensure safety when working with high voltage parts of the hybrid electric vehicles because rubber is an insulator i.e. it does not allow direct contact of electricity with the body. These gloves are really important because these gloves can protect the technicians from electric shocks servicing the high voltage system on hybrid electric vehicles. It needs to be ensured that the gloves are of pure rubber and have no holes or cuts to ensure complete insulation. These gloves are also called class “0” gloves or gloves rated at 1000 V are used for this.

Similarly there are methods to shut off the high voltage to components and circuits in hybrid cars. These methods include but are not limited to:

• Turning the ignition off

• Disconnecting the 12-volt auxiliary battery

• Removing the main fuse, relay, or HV plug

So both the technicians are correct

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A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
3 years ago
One of the element carbon combines with one of the element oxygen to form one of the compound carbon dioxide.
saul85 [17]
False. C + O --> CO not CO2. Carbonmonoxide
3 0
3 years ago
Read 2 more answers
A person weighing 785 newtons on the surface of Earth would weigh 298 newtons on the surface of Mars.
rodikova [14]

Answer:

The gravitational field strength on the surface of Mars = 3.72 m/s²

Explanation:

Gravitational Field Strength: This can be defined as the force per unit mass which is exerted at that point. its direction is the force exerted on a mass in a gravitational field. The S.I unit of gravitational field strength is m/s²

Mathematically, Gravitational field is represented as,

g = F/m ..................... Equation 1.

m = F/g ..................... Equation 2.

Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.

From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

m = 785/9.8

m = 80.10 kg.

On the surface of Mars.

Given: m = 80.10 kg, F = 298.

Substituting into equation 2

g = 298/80.1

g = 3.72 m/s²

Thus the gravitational field strength on the surface of Mars = 3.72 m/s²

3 0
4 years ago
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.146 kg baseball crashing through the pane
Nonamiya [84]

Answer:

Impulse, |J| = 0.6716 kg-m/s

Force, F = 63.35 N

Explanation:

It is given that,

Mass of the baseball, m = 0.146 kg

Initial speed of the ball, u = 15.3 m/s

Final speed of the ball, v = 10.7 m/s

To find,

(a) The magnitude of this impulse.

(b) The magnitude of the average force of the glass on the ball.

Solution,

(a) Impulse of an object is equal to the change in its momentum. It is given by :

J=m(v-u)

J=0.146\ kg(10.7-15.3)\ m/s

J = -0.6716 kg-m/s

or

|J| = 0.6716 kg-m/s

(b) Another definition of impulse is given by the product of force and time of contact.

t = 0.0106 s

J=F\times \Delta t

F=\dfrac{J}{\Delta t}

F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}

F = 63.35 N

Hence, this is the required solution.

6 0
3 years ago
A 30 kg dog runs at a speed of 15 m/s. What is the dog's kinetic energy (in jewels)? Round the answer to two significant digits.
mel-nik [20]

Answer:

It would be 12.34

Explanation:

6 0
3 years ago
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