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Masja [62]
4 years ago
9

Did you think about this over Christmas? I did (-: Before Christmas a 65kg student consumes 2500 Cal each day and stays at the s

ame weight. For three days in a row while visiting her parents she eats 3500 Cal and, wanting to keep from gaining weight decides to "work off" the excess by jumping up and down at the Christmas tree. With each jump she accelerates to a speed of 3.2 m/s before leaving the ground. a) How high will she jump each jump? b) How many jumps must she do to keep her weight? Assume that the efficiency of the body in using energy is 25%. c) Do you suggest that is a reasonable way for the student not to gain weight over Christmas? d) Possible enhancement: What other way/ways would you suggest for the student to keep her weight?
Physics
1 answer:
Gnesinka [82]4 years ago
7 0

Answer:

a)  Em = 332.8 J , b) # jump = 13, c)   It is reasonable since there are not too many jumps , d) lower the calories consumed

Explanation:

a) Let's use energy conservation

Initial. On the floor

             Em₀ = K = ½ m v²

Final. The highest point

             Emf = U = m g h

Energy is conserved

             Em₀ = Emf

             ½ m v² = m g h

             h = ½ v² / g

            h = ½ 3.2² /9.8

            h = 0.52 m

b) When he was at home he maintained his weight with 2500 cal / day. In his parents' house he consumes 3500 cal / day, the excess of calories is

            Q = 3500 -2500 = 1000cal / day

Let's reduce this value to the SI system

             Q = 1000 cal (4,184 J / 1 cal) = 4186 J / day

Now the energy in each jump is

               Em = K = ½ m v²

               Em = ½ 65 3.2²

               Em = 332.8 J

They indicate that the body can only use 25% of this energy

              Em effec = 0.25 332.8 J

              Em effec = 83.2 J

This is the energy that burns the body

Let's use a Proportion Rule (rule of three), if a jump spends 83.2J how much jump it needs to spend 1046 J

              # jump = 1046 J (1 jump / 83.2 J)

              # jump = 12.6 jumps / day

              # jump = 13  

c) It is reasonable since there are not too many jumps

d) That some days consume more vegetables to lower the calories consumed

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Masja [62]

Answer:

speed of the charge electric is  v = - (Eo q/m) cos t

Explanation:

The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,

          F = q Eo sint

a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity

        F = ma

        q Eo sint = ma

        a = Eo q / m sint

        a = dv / dt

        dv = adt

        ∫ dv = ∫ a dt

        v-vo = I (Eoq / m) sin  t dt

        v- vo = Eo q / m (-cos t)

We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0

        v = - (Eo q / m) cos t

b) Kinetic energy

       

         K = ½ m v2

          K = ½ m (Eoq / m)²2 (sint)²

         K = ¹/₂  Eo² q² / m sin² t

c) The average kinetic energy over a period

          K = ½ m v2

         <v2> = (Eoq / m) 2 <cos2 t>

The average of cos2 t = ½, substitute and calculate

          K = ½ m (Eoq / m)²  ½

          K = ¼ Eo² q² / m

7 0
4 years ago
Substances that move to the stronger parts of a magnetic field are termed ______ substances; the atomic feature responsible for
Arisa [49]

Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

<h3>What is a paramagnetic substance?</h3>

A paramagnetic substance is the substance that possess unpaired electrons that are heavily attracted in a magnetic field.

A magnetic field is defined as the field that exists around a magnet that produces a field of force.

Examples of paramagnetic substance include the following:

  • aluminum,
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  • copper.
  • Chromium, and
  • Manganese.

These substances are known as paramagnetic substances because they possess a high number of unpaired electrons.

Other properties of a paramagnetic substance include the following:

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Therefore, Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

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3 0
1 year ago
A shallow groove on the surface of the cortex is called a ________.
Juli2301 [7.4K]

Answer:

sulcus

Explanation:

A sulcus is an indentation or depression in the brain that causes it to look like it  ridges or folds

Cerebral sulci and fissures are grooves between the adjacent gyri on the surface of the cerebral hemispheres.

Sulci can be basically can be divided into three basic function

limiting sulcus: This happens to  develop between areas differing in structure and function, for example central sulcus

axial sulcus: This develops along the axis of a rapidly growing/developing area (e.g. calcarine sulcus)

operculated sulcus: a sulcus may be between two structurally-different areas and a third sulcus may lie in its wall and does not appear on the surface (e.g. lunate sulcus)

3 0
3 years ago
Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs.
Step2247 [10]

Answer:

Part a)

F = 15600 N

Part b)

force is 21.2 times more than the weight of the person

Explanation:

Part a)

As it is given that the distance after which he stopped is given as

d = 1.50 cm

here finally it stops so final speed is given as

v_f = 0

initial speed is given as

v_i = 6 m/s

now by the equation of kinematics we know that

v_f^2 - v_i^2 = 2a d

0^2 - 6^2 = 2(a)(0.015)

a = -1200 m/s^2

Now the force on the leg is given as

F = ma

m = mass of leg = 13 kg

F = (13 kg)(-1200 m/s^2)

F = 15600 N

Part b)

Force due to weight of the object

F_g = Mg

here we know

M = 75.0 kg

F_g = 75(9.81)

F_g = 735.75 N

Now we know that the ratio of the weight with the force on leg is given as

\frac{F}{F_g} = \frac{15600}{735.75} = 21.2 N

so force is 21.2 times more than the weight of the person

5 0
4 years ago
Physics Question: Determine the work that is being done by tension in pulling the box 193.0 cm along the table. Also determine t
lozanna [386]

1) The work done by tension is 12.0 J

2) The final speed of the box is 1.57 m/s

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

F = 6.2 N is the force applied (the tension in the string)

d = 193.0 cm = 1.93 m is the displacement

\theta=0 assuming that the string is horizontal

Substituting, we find the work done by the tension:

W=(6.2)(1.93)(cos 0)=12.0 J

2)

In order to determine the final speed of the box, we need to determine its acceleration first.

Beside the tension, acting forward, the other force acting horizontally on the box is the force of friction, whose magnitude is

F_f = \mu mg

where

\mu=0.16 is the coefficient of friction

m = 2.8 kg is the mass of the box

g=9.8 m/s^2 is the acceleration of gravity

Substituting,

F_f = (0.16)(2.8)(9.8)=4.4 N

Therefore, the net force on the box is

F=6.2  N - 4.4 N = 1.8 N

And the acceleration can be found by using Newton's second law:

F=ma

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F = 1.8 N is the net force

m = 2.8 kg is the mass of the box

a is the acceleration

Solving for a,

a=\frac{F}{m}=\frac{1.8}{2.8}=0.64 m/s^2

Now we can finally find the final speed using the suvat equation:

v^2-u^2=2as

where

v is the final speed

u = 0 is the initial speed

a=0.64 m/s^2 is the acceleration

s = 1.93 m is the displacement

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(0.64)(1.93)}=1.57 m/s

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3 0
4 years ago
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