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Deffense [45]
3 years ago
9

Technician A says that the starter motor used to crank diesel engines can draw up to 400 amps of current. Technician B says that

high resistance on the insulated side of a starter motor circuit would cause higher than specified starter motor current draw. Who is correct?a. Technician A
b. Technician B
c. Both Technician A and Technician B
d. Neither Technician A nor Technician B
Physics
1 answer:
Aleks04 [339]3 years ago
8 0

Answer: Option A : Technician A

Explanation:

The statement/observation, "that the starter motor used to crank diesel engines can draw up to 400 amps of current" made by Technician A is correct.

A diesel engine uses up to 400+ Amperes of electricity to start up a diesel engine in the ignition chamber of motor engine.

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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
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Describe the tide requirements necessary to create a tidal power plant.
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Answer:

La energía mareomotriz se produce gracias al movimiento generado por las mareas, esta energía es aprovechada por turbinas, las cuales a su vez mueven la mecánica de un alternador que genera energía eléctrica, finalmente este último esta conectado con una central en tierra que distribuye la energía hacia la comunidad.    

6 0
2 years ago
Which best explains how the diffraction pattern observed in young’s experiment supports the wave theory of light?.
KIM [24]

Answer: the airy pattern can only arise from wave propagation

Explanation:if particles went in straight lines through a slit, they would progate linearly and not interfere. The airy pattern arises from diffraction as waves interfere, producing peaks (constructive interference where peaks of waves from each slit coincide) and troughs (destructive interference where peaks and troughs of waves from each slit cancel out). If intensity rather than field is measured nodes occur where 0 values line up instead of troughs

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2 years ago
Distance travelled/time taken gives?
skad [1K]

Answer:

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3 years ago
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Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam
sasho [114]
Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
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3 years ago
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