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2 years ago

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john doe operates a crane that can pick up 3.0 tons of excavated earth in an hour. John's wages are $35 per hour. What, then, is

the cost of picking up 85kg of excavated earth?

1 answer:

levacccp [35]2 years ago

8 0

Answer:

1

Explanation:

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A chain link fence should be cut quickly with a

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Answer: it should be cut with a chainsaw

Explanation:

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2 years ago

The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to tra

Olin [163]

Answer:

t = 0.196 s

Explanation:

The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement

v = x / t

t = x / v

calculate

t = 2/102

t = 0.196 s

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2 years ago

Which of the following choices is an example of an allele?

VikaD [51]

Answer:

Blue eye-color gene

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2 years ago

What does it mean when work is positive?

jenyasd209 [6]

Both positive work and negative work have meaning: Positive work follows when the force has a component parallel to the displacement. Positive work adds energy to a system. Negative work follows when the force has a component opposite or against the displacement.

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2 years ago

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res

murzikaleks [220]

Answer:

(a) $V=11.86\ V$

(b) $V=9.76\ V$

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

$V=R.I$

(a) The electromagnetic force of the battery is $\varepsilon =12.6\ V$ and its internal resistance is $R_i=0.06\ \Omega$ . Knowing the equivalent resistance of the headlights is $R_e=5.2\ \Omega$ , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

$\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)$

Solving for i

$\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A$

i=2.28\ A

The potential difference across the headlight bulbs is

$V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V$

$V=11.86\ V$

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

$V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V$

5 0

2 years ago