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2 years ago

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The food calorie, equal to 4186J , is a measure of how much energy is released when food is metabolized by the body. A certain b

rand of fruit-and-cereal bar contains 140 food calories per bar.If a 69.0kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
If, as is typical, only 20.0% of the food calories go into mechanical energy, what would be the answer to part (a)?

1 answer:

vovangra [49]2 years ago

5 0

<h2>The hiker will go up to 850 m on the hill</h2>

Explanation:

The total energy gained by the hiker = 140 x 4186 J

This energy is consumed in the potential energy acquired , while climbing up the hill.

The potential energy P.E = mass of hiker x acceleration due to gravity x height

Thus

140 x 4186 = 69 x 10 x h

or h = $\frac{4186x140}{69x10}$ = 850 m

If the 20% of the total energy is used

the height h₀ = $\frac{0.2x4186x140}{69x10}$ = 170 m

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A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener

Bingel [31]

The ball was dropped from a height 20 meters

Explanation:

The given is

1. A ball is dropped from the top of a cliff

2. By the time it reaches the ground, all the energy in its gravitational

potential energy store has been transferred into its kinetic energy

store, that mean K.E = P.E

3. The ball is travelling at 20 m/s when it hits the ground

4. The gravitational field strength is 10 N/kg

We need to find the height that the ball dropped from it

The ball dropped from the top of a cliff means the initial speed is 0

→ K.E = $\frac{1}{2}m(v^{2}-v_{0}^{2})$

where v is the final speed, $v_{0}$ in the initial speed and m

is the mass

→ v = 20 m/s and $v_{0}$ = 0 m/s

→ K.E = $\frac{1}{2}m(20^{2}-0^{2})$

→ K.E = $\frac{1}{2}m(400)$

→ K.E = 200 m joules ⇒ when the ball hits the ground

→ P.E = m g h

where g is the gravitational field strength, m is the mass and h is

the height

→ g = 10 N/kg

→ P.E = m(10)(h)

→ P.E = 10 m h joules

→ P.E = K.E

→ 10 m h = 200 m

Divide both sides by 10 m

→ h = 20 meters

The ball was dropped from a height 20 meters

Learn more

You can learn more about gravitational potential energy in brainly.com/question/1198647

#LearnwithBrainly

8 0

1 year ago

A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which

artcher [175]

Answer:

h = 10 m

Explanation:

given,

mass of platform = 50 Kg

Kinetic energy = 5000 J

height from which the diver dove = ?

taking acceleration due to gravity = 10 m/s²

using conservation of energy

Kinetic energy is converted into mechanical energy

K.E = P.E

K.E = m g h

5000 = 50 x 10 x h

500 h = 5000

$h = \dfrac{5000}{500}$

h = 10 m

The height from which the diver dove is equal to h = 10 m

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1 year ago

The bottom of the inner curve of a hook is called

padilas [110]

Slip hook or blunt hook

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2 years ago

A car traveling at 91.0 km/h approaches the turn off for a restaurant 30.0 m ahead. If the driver slams on the brakes with the a

eduard

Answer: 49.92 m

Explanation:

In this situation the following equation will be useful:

$V^{2}=V_{o}^{2} +2 a d$

Where:

$V=0 m/s$ is the final velocity of the car, when it finally stops

$V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s$ is the initial velocity of the car

$a=-6.4 m/s^{2}$ is the constant acceleration of the car after the driver slams on the brakes

$d$ is the stopping distance

Isolating $d$ :

$d=\frac{-V_{o}^{2}}{2a}$

$d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}$

$d=41.919 m \approx 41.92 m$

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2 years ago

Have you ever chewed on a wintergreen mint in front of a mirror in the dark? If you have, you may have noticed some sparks of li

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Answer:

Part a)

$E = 3.66 eV$

Part b)

$\lambda = 508.5 nm$

Explanation:

Part a)

change in the energy due to decay of photon is given as

$E = h\nu$

here we know that

$\nu = 8.88 \times 10^{14} Hz$

now we have

$E = (6.6 \times 10^{-34})(8.88 \times 10^{14})$

$E = 5.86 \times 10^{-19} J$

$E = 3.66 eV$

Part b)

While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy

so we have

$\Delta E = \frac{2}{3}(3.66 eV)$

$\Delta E = 2.44 eV$

now to find the wavelength we have

$\Delta E = \frac{hc}{\lambda}$

$2.44 = \frac{1242}{\lambda}$

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2 years ago