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GuDViN [60]
3 years ago
5

Technician A says that to perform a maximum output test on a typical heavy-duty truck charging circuit, the engine should be run

at 1500 rpm. Technician B says that alternator rpm is typically at least three times engine rpm. Who is correct?
Physics
1 answer:
Temka [501]3 years ago
3 0

Answer:

Both Technicaian A and B is correct

Explanation:

An alternator is a device that converts mechanical energy into electrical energy. In every alternator, there is unavoidable power losses. Such losses could be mechanical,drive belt or getting the alternator's bearing heated and electrical loss.more also changing magnetic field also causes some losses in an alternator. The diodes of an alternator get hot when there is drop in votage.At reasonable maximum engine speeds, For instance when it read 1500 RPM to redline and there is heavy loads, starting from the idle through the red line with light electrical accessory loads, Nothing will happen to battery, it will just goes along for the ride. Unless it wait for the alternator to fall below operating speeds.

Therefore, when technician A says that to perform a maximum output test on a typical heavy-duty truck charging circuit, the engine should be run at 1500 rpm, and Technician B says that alternator rpm is typically at least three times engine rpm. Both of them is is correct.

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For general projectile motion, the horizontal component of a projectile's velocity aSelect one:O a. continuously increases.O b.
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d. remains a non-zero constant.

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3 years ago
A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are rele
ZanzabumX [31]

Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

u^2-u'^2 = 2as

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

u^2-0^2 = 2\times9.8\tiimes1.08

or

u = \sqrt{21.168}

or

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed (v_f) = v = 4.6 m/s

Then conservation of energy says  

\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2  

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s  

u₂ = velocity of the basketball before collision= 4.6 m/s  

v₁ =  velocity of the tennis ball after collision  

v₂ = velocity of the basketball  after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2

substituting the values in the above equation we get

v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6

or

v = -8.565-3.965

or

v = -12.53m/s

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

K.E = \frac{1}{2}mv^2

or

K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m

5 0
3 years ago
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