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Norma-Jean [14]
3 years ago
15

Professor Goetz took a heavy marine 100 foot, 0.75 lb per foot rope and through it through his office window. One end of the rop

e touches the ground. Professor Goetz missed his lunch and can only exert 300kcal (the equivalent of a healthy sandwich). He realized he does not know if he has enough energy to lift the rope back into his office. a) How much energy in kcalories is needed to lift the rope half way from the ground. b) Calculate whether he is capable of lifting the rope all the way up. c) A person who eat a candy gains extra 50kcal. If this energy is used to lift the rope, how many feet will it be lifted from the ground.
Physics
1 answer:
trasher [3.6K]3 years ago
6 0

Answer:

a. E = 11.90876 kcal

b. E = 23.81753 k cal

c. d = 209.93 ft

Explanation:

Knowing the relation between the energy and the food to realize how may energy do you need in each situation

a.

h = 100 ft / 2 = 50 ft

We = 0.75 lb * 100 * 9.8 = 735 J

E = We * h = 735 J * 50 ft = 36750 J

E to Kcal = 36750 J  ⇒   11.90876 k cal

E = 11.90876 kcal

b.

E = 735 * 100 ft = 73500 J

E to Kcal  73500 J  ⇒  23.81753 k cal

E = 23.81753 k cal

c.

Now to determine the distance with a candy of 50 kcal

d = 100 ft   ⇒   23. 81753 k cal

        d        ⇒   50 k cal  

d = ( 100 * 50 )  /  23.81753

d = 209.93 ft

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The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

h = v_0_yt + \frac{1}{2} gt^2

where;

  • <em>t is the time of motion </em>
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h = \frac{1}{2} gt^2

The time taken by the first stone to hit the ground is calculated as;

t_1 = \sqrt{\frac{2h}{g} }

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

t_2 = \sqrt{\frac{2h}{g} } + 1

t_2 = t_1 + 1

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"<em>Your question is not complete, it seems be missing the following information;"</em>

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0
2 years ago
An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef
Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field B=5.0\times10^{-5}\ T

(a). We need to calculate the induced emf

Using formula of emf

E = Blv\sin\theta

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}

E=1.398\times10^{-3}\ V

We need to calculate the induced current

E =IR

I=\dfrac{E}{R}

Put the value into the formula

I=\dfrac{1.398\times10^{-3}}{3.0}

I=0.5\ mA

(b). We need to calculate the power dissipated through the resistor

Using formula of power

P=I^2 R

Put the value into the formula

P=(0.5\times10^{-3})^2\times3.0

P=7.5\times10^{-7}\ Watt

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

6 0
3 years ago
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True

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3 0
3 years ago
usted / el partido de fútbol Usted va al partido de fútbol. Question 1 with 1 blankyo / la biblioteca Question 2 with 1 blanknos
34kurt

Answer:

1- Yo VOY A la biblioteca

2- Nosotros VAMOS A la piscina

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4- Tú VAS de excursión

5- Ustedes VAN AL parque municipal

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Explanation:

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