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Vedmedyk [2.9K]
3 years ago
6

The food calorie, equal to 4186 J, 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 150 food calories per bar. Part A If a 68.0 kg 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?
Physics
1 answer:
rewona [7]3 years ago
5 0

Answer:

The hiker has to climb 941.26 to ''work off'' the calories.

Explanation:

Let's first find out how much energy the hiker gets from 150 food calories.

This is:

Energy = 150 * 4186

Energy = 627.9 kJ

To burn all of this off, we set it equal to the gravitational potential energy, and then solve for the height.

This is:

Gravitational Potential Energy (G.P.E) = 629700

G.P.E = mass * gravity * height

Thus,

mass * gravity * height = 629700

68 * 9.81 * height = 629700

height = 941.26 meters

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WINSTONCH [101]

Answer:

knee to chest

Explanation:

I took the test

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2 years ago
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How to find the magnitude and direction of a resultant velocity?
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Find the horizontal components vcos30 ...one goes right and one goes left so they cancel each other.
Find vertical components vsin30.....there are two of them.... so 2vcos30....hey presto... resultant velocity = 2vCos30
5 0
3 years ago
A soccer player icks a rock horizontally off a 40m high cliff into a pool f water if the player hears the sound of the splash s
Semenov [28]

Answer:

v = 9.936 m/s

Explanation:

given,

height of cliff = 40 m

speed of sound = 343 m/s

assuming that time to reach the sound to the player = 3 s

now,

time taken to fall of ball

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 40}{9.8}}

t = 2.857 s

distance

d = v  x t

d = v x 2.875

time traveled by the sound before reaching the player

t_0 = t - t_{fall}

t_0 = 3 - 2.875

t_0 = 0.143 s

distance traveled by the wave in this time'

r = 0.143 x 343

r= 49.05 m

now,

we know.

d² + h² = r²

d² + 40² = 49.05²

d =28.387 m

v x 2.875=28.387 m

v = 9.936 m/s

7 0
3 years ago
The force an ideal spring exerts on an object is given by Fx = -kx, where x measures the displacement of the object from its equ
shusha [124]

Answer:

The work done by this force can be found via the following formula

W = \int{F(x)} \, dx = \int\limits^0_{-20} {(-kx)} \, dx = \frac{-kx^2}{2}\left \{ {{x=0} \atop {x=-20}} \right. = \frac{-60*(-20)^2}{2} \\W = -12000J

Explanation:

Alternatively, the work done by the object is equal to the elastic potantial energy done by the spring.

U = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 =0 - \frac{1}{2}60(-20)^2 = -12000J

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3 years ago
The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO.
Genrish500 [490]

Answer:

Explanation:

Given a square side loop of length 10cm

L=10cm=0.1m

Then, Area=L²

Area=0.1²

Area=0.01m²

Given that, frequency=60Hz

And magnetic field B=0.8T

a. Flux Φ

Flux is given as

Φ=BA Sin(wt)

w=2πf

Φ=BA Sin(2πft)

Φ=0.8×0.01 Sin(2×π×60t)

Φ=0.008Sin(120πt) Weber

b. EMF in loop

Emf is given as

EMF= -N dΦ/dt

Where N is number of turns

Φ=0.008Sin(120πt)

dΦ/dt= 0.008×120Cos(120πt)

dΦ/dt= 0.96Cos(120πt)

Emf=-NdΦ/dt

Emf=-0.96NCos(120πt). Volts

c. Current induced for a resistance of 1ohms

From ohms law, V=iR

Therefore, Emf=iR

i=EMF/R

i=-0.96NCos(120πt) / 1

i=-0.96NCos(120πt) Ampere

d. Power delivered to the loop

Power is given as

P=IV

P=-0.96NCos(120πt)•-0.96NCos(120πt)

P=0.92N²Cos²(120πt) Watt

e. Torque

Torque is given as

τ=iL²B

τ=-0.96NCos(120πt)•0.1²×0.8

τ=-0.00768NCos(120πt) Nm

8 0
3 years ago
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