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3 years ago

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Technician A says that the battery must be in good condition and at least 75% charged to accurately test an alternator. Technici

an B says that alternator output should be tested at idle. Which technician is correct?a. technician A.
b. technician B.
c. both A and B.
d. neither A nor B.

1 answer:

aalyn [17]3 years ago

5 0

Answer:Technician A

Explanation:

Technician A statement is correct as

The battery is required to start the vehicle which, in effect, rotates the alternator at sufficient speed to keep the battery charged. This means if the battery is low it is not possible to start the vehicle and thus we are unable to test the alternator.

That is the battery is pre-requisite to test the alternator. So the battery must be at least a 75 % charge to test the alternator.

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Answer:

CHEMICAL WHETEARING

Explanation:

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3 years ago

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A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo

Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

$E_{1}=E_{pot}+E_{kin}+E_{elas}\\$

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

$E_{2}=E_{kin}+E_{pot}$

Now we can use the first statement to get the first equation:

$E_{1}+W_{1-2}=E_{2}$

where:

W₁₋₂ = work from the state 1 to 2.

$E_{k}=\frac{1}{2} *m*v^{2} \\$

$E_{pot}=m*g*h$

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

$70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5$

$58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]$

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3 years ago

A pillow with mass of 0.3 kg sits on a bed with a coefficient of static friction of 0.6. What is the maximum force of static fri

lys-0071 [83]

The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body.

P = W = mass x acceleration due to gravity

P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N

Solving for the static friction force (F),
F = P x c

F = (2.94 N) x 0.6 = 1.794 N

Therefore, the maximum force of static friction is 1.794 N.

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3 years ago

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Two wheel gears are connected by a chain. The larger gear has a radius of 8 centimeters and the smaller gear has a radius of 3 c

dezoksy [38]

Answer:

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

Explanation:

As we know that the small gear completes 24 revolutions in 20 seconds

so the angular speed of the smaller gear is given as

$\omega = 2\pi\frac{24}{20}$

$\omega = 2.4\pi rad/s$

Now we know that the tangential speed of the chain is given as

$v = r \omega$

so we have

$v = (3 cm)(2.4\pi)$

$v = 7.2 \pi cm/s$

$v = 432\pi cm/min$

Since both gears are connected by same chain so both have same linear speed and hence correct answer will be

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

8 0

3 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0

3 years ago