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2 years ago

Most farm equipment that attaches to a tractor is powered by its own independent motor.

Engineering

1 answer:

Pavlova-9 [17]2 years ago

3 0

Answer:

False

Explanation:

its self explanatory. a tracter pulls the equipment, so why power the equipment? the tracter is giving it free energy.

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Question: 10 of 15

Anvisha [2.4K]

Answer:

Leg length

Explanation:

The distances from the root to the edges of the legs (toes) and the height of the crown are basic measurements.

3 0

2 years ago

Which statement about lean manufacturing is true when you compare it to mass production?

Len [333]

Where are the statements then bbs lol

6 0

2 years ago

If block A of the pulley system is moving downward at 6 ft>s while block C is moving down at 18 ft>s, determine the relati

kramer

Answer:

Explanation:

The detailed steps and appropriate calculation with analysis is as shown in the attachment.

3 0

2 years ago

You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19

Pepsi [2]

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

$V = A*t$

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

$V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}$

Now, using the density we can find the mass:

$m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g$

Finally, with the Avogadros number ( $N_{A}$ ) and with the atomic mass (A) we can find the number of atoms (N):

$N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms$

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0

2 years ago

The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery

Lynna [10]

Answer:

capacity = 0.555 mAh

capacity = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C

capacity = 2 × ampere × second ...............1

put here value and we get

and 1 Ah = 3600 C

capacity = $\frac{2}{3600}$

capacity = 0.555 mAh

and

when it is in parallel than capacity will be

capacity = Q1 +Q2 ...............2

capacity = 1800 + 1800

capacity = 3600 mAh

3 0

3 years ago