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hram777 [196]
1 year ago
6

When you cook a marshmallow on a metal poker tool over an open flame, energy is transferred. Identify the three different ways t

hermal energy is transferred in this situation. Be sure to name the specific places or phenomenon in this system where each of the types of thermal energy transfer is occurring and the direction the energy is being transferred (from what to what).
Physics
1 answer:
ss7ja [257]1 year ago
3 0

When we cook a marshmallow on a metal poker tool over an open flame, there are three ways in which heat energy is transferred: Conduction, convection, and radiation.

<h3>Heat energy transfer</h3>

Heat transfer is the natural transfer of heat from an object with a higher temperature to an object with a lower temperature. Heat transfer can occur in three ways, namely conduction, convection, and  radiation.

  1. Conduction occurs when heat flows from a place with a high temperature to a place with a lower temperature using a fixed heat-conducting medium. Heat transfer from the open flame to the marshmallows via direct fire contact with the marshmallows is an example of conduction.
  2. Convection is the transfer of heat by means of a stream in which the intermediate substance also moves. If the particles move and cause heat to propagate, convection will occur. The hot air rising from the flames burning the marshmallows is an example of convection.
  3. Radiation is heat transfer without a medium. Radiation can also usually be accompanied by light. The direct transfer of heat from the flame to the marshmallow in the form of waves is an example of radiation.

Learn more about heat transfer here: brainly.com/question/16055406

#SPJ4

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A chemically broken down into sugar
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3 years ago
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Anvisha [2.4K]

The speed of the rock at 20 m is 34.3 m/s

Explanation:

We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy

K_i is the initial kinetic energy

U_f is the final potential energy

K_f is the final kinetic energy

The equation can also be rewritten as  follows:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 100 kg is the mass of the rock

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

h_i = 80 is the initial height

u = 0 is the initial speed  (the rock starts at rest)

h_f = 20 m is the final height of the rock

v is the final speed when h = 20 m

And solving for v, we find:

v=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(80-20)}=34.3 m/s

Learn more about kinetic energy and potential energy here:

brainly.com/question/6536722

brainly.com/question/1198647  

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#LearnwithBrainly

5 0
3 years ago
A hard-boiled egg of mass 46.0 gg moves on the end of a spring with force constant 25.6 N/mN/m . The egg is released from rest a
soldi70 [24.7K]

Answer:

0.022kg/s

Explanation:

We are given that

Mass of boiled egg=46 g=\frac{46}{1000} kg=0.046 kg

1kg=1000 g

Constant force=F=25.6 N/m

Initial displacement=A_1=0.296 m

Final displacement=A_2=0.12 m

Time=t=4.55 s

Damping force=F_x=-bv_x

We have to find the  magnitude of damping constant b.

We know that the displacement of the oscillator under damping motion is given by

x=Ae^{-\frac{b}{2m}t}cos(w't+\phi)

For maximum displacement cos(w't+\phi)=1

Therefore , x=A_2

Substitute the values

A_2=A_1e^{-\frac{-b}{2m}t}

e^{-\frac{b}{2m}t}=\frac{A_2}{A_1}

-\frac{b}{2m}t=ln\frac{A_2}{A_1}

lnx=y\implies x=e^y

Substitute the values

-\frac{b}{2\times 0.046}\times 4.55=ln\frac{0.12}{0.296}

\frac{2\times 0.046}{4.55b}=ln\frac{0.296}{0.12}

\frac{2\times 0.046}{4.55}=0.9b

b=\frac{2\times 0.46}{4.55\times 0.9}=0.022kg/s

Hence,the  magnitude of damping constant b=0.022kg/s

3 0
3 years ago
A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5
svlad2 [7]
The magnetic force experienced by the proton is given by
F=qvB \sin \theta
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and \theta the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so \sin \theta=1 and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
F=ma

So we have
ma=qvB
from which we can find the magnitude of the field:
B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T
4 0
3 years ago
A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87° from the direction of th
77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, v=8.2\times 10^3\ m/s

Angle between velocity and the magnetic field, \theta=87

Charge, q=5.7\ \mu C=5.7\times 10^{-6}\ C

Magnetic force, F = 0.002 N

The magnetic force is given by :

F=qvB\ sin\theta

B is the magnetic field  

B=\dfrac{F}{qv\ sin\theta}

B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

5 0
3 years ago
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