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jolli1 [7]
4 years ago
8

The nutritional calorie (Calorie) is equivalent to 1 kcal. One pound of body fat is equivalent to about 4.1 x 103 Calories. If a

runner expends 1850 kJ/h, how long (h) does she have to run to work off 7.56 lb of body fat? Enter your answer in decimal format with one decimal place.
Physics
1 answer:
Luba_88 [7]4 years ago
5 0

Answer:

t = 70.13 hours

Explanation:

As we know that 1 lb fat can be used to expend 4.1 k Calorie of energy

so here we know that we have to use 7.56 lb of body fat

so here the total energy that is to be utilized is given as

Q = (4.1 \times 10^3)(7.56)

so we will have

Q = 30996 Calorie

now we know that runner expends the energy at rate of 1850 kJ/h

so we have

E = 1850 \times 10^3 kJ/h

so it is

E = 0.44 \times 10^3 calorie/h

now we know that time to burn this energy is given as

t = \frac{Q}{E}

t = \frac{30996}{0.44 \times 10^3}

t = 70.13 hours

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A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a
Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

F_N = F_C - F_G = m\frac{v^{2} }{r} - mg

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

E_{tot}=mgh + \frac{1}{2} mv^{2}

At the top of the hill:

E_{tot}= mgh_{hill}

At the top of the loop:

E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}

Setting both energies equal and canceling the mass m gives:

gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}

Solving for v:

v^{2} = 2g(h_{hill}-h_{loo}_p)

Using v in the first equation:

F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg

F_N = 955.5N

7 0
3 years ago
_____ motion occurs when tectonic plates move apart from each other. A. Divergent B. Transform C. Convergent
Likurg_2 [28]
A. Divergent (pretty sure)
4 0
4 years ago
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A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene
Yuri [45]

Explanation:

The given data is as follows.

            U_{a} = 5.4 \times 10^{-8} J

        W_{/text{a to b}} = -1.9 \times 10^{-8} J

        Electric potential energy (U_{b}) = ?

Formula to calculate electric potential energy is as follows.

            U_{b} = U_{a} - W_{/text{a to b}}

                        = 5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)

                        = 7.3 \times 10^{-8} J

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is 7.3 \times 10^{-8} J.

6 0
4 years ago
A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.
mash [69]

a) The acceleration of gravity is 4.96 m/s^2

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2

b)

The critical velocity for a satellite orbiting around a planet is given by

v=\sqrt{\frac{GM}{R+h}}

where we have again:

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s

Converting into km/h,

v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

T=\frac{2\pi (R+h)}{v}

where

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

v = 6668 m/s

Substituting,

T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s

d)

One day consists of:

t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

n=\frac{t}{T}=\frac{86400}{8452}=10.2

e)

The escape velocity for an object in the gravitational field of a planet is given by

v=\sqrt{\frac{2GM}{R+h}}

where here we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

Substituting, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s

f)

We can apply again the formula to find the escape velocity for the rocket:

v=\sqrt{\frac{2GM}{R+h}}

Where this time we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=0, because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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Simply be used as a reference point <span>to describe its position. a fact forming the basis of an evaluation or assessment; criterion. They had few cultural </span>reference points<span> in common.</span>
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