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3 years ago

A student is investigating the best method for heating marshmallows using an open flame.

Physics

1 answer:

Llana [10]3 years ago

7 0

Answer:

The answer is Conduction.

Explanation:

In the first scenario given in the question where the student is holding the marshmallows on the side of the flame, she is subjecting the marshmallow the a heat transfer by radiation since the marshmallow is not in direct contact with the fire.

Heat transfer by radiation is done so with the heating of the air between the object and the heat source and does not need any contact between the source and the heated material.

And later when she places the marshmallow over the flames, the heat transfer type she is observing is heat transfer by conduction.

I hope this answer helps.

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A 35 kg child slides down a playground slide at a constant speed. The slide has a height of 3.8 m and is 8.0 m long. Find the ma

ad-work [718]

Answer:

The magnitude of the kinetic frictional force acting on the child is 162.93 N

Explanation:

Given;

mass of the child, m = 35 kg

height of the slide, h = 3.8 m

length of the slide, d = 8.0 m

The change in thermal energy associated with the kinetic frictional force is calculated as follows;

$\Delta E_{th} + \Delta K.E + \Delta U = 0\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i) =0\\\\since \ the \ speed \ is \ constant, \ v_f = v_i \ and \ \Delta K.E = 0\\\\Also, \ final \ height \ , h _f= 0\\\\\Delta E_{th} - mgh_i = 0\\\\\Delta E_{th} = mgh_i\\\\\Delta E_{th} = 35 \times9.8 \times 3.8\\\\\Delta E_{th} = 1303.4 \ J$

The magnitude of the kinetic frictional force that produced this thermal energy is calculated from the work done by frictional force;

$\Delta E_{th} = F \times d\\\\F = \frac{\Delta E_{th} }{d} \\\\F = \frac{1303.4}{8} \\\\F = 162.93 \ N$

Therefore, the magnitude of the kinetic frictional force acting on the child is 162.93 N

8 0

3 years ago

A 0.50-kg toy is attached to the end of a 1.0-m very light string. The toy is whirled in a horizontal circular path on a frictio

Deffense [45]

Answer:

26.5 m/s

Explanation:

The tension in the string provides the centripetal force that keeps the toy in circular motion. So we can write:

$T=m\frac{v^2}{r}$

where:

T is the tension in the spring

m = 0.50 kg is the mass of the toy

r = 1.0 m is the radius of the circle (the length of the string)

v is the speed of the toy

The maximum tension in the string is

T = 350 N

If we substitute this value into the equation, we find the maximum speed that the mass can have before the string breaks:

$v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(350)(1.0)}{0.50}}=26.5 m/s$

3 0

3 years ago

Titanium is a metal used to make golf clubs. A rectangular bar of this metal measuring 1.96 cm x 2.19 cm x 2.63 cm was found to

masha68 [24]

$4.012\frac{gr}{cm^3}$

Explanation

the density of an object is given by:

$\text{Density(d)}=\frac{mass(m)}{\text{volume(v)}}$

Step 1

find the volume of the bar

a)find the volume of the rectangular bar.

the volume of a rectangular prism is given by:

$\text{Volume}=\text{ length}\cdot widht\cdot depth$

replace

$\begin{gathered} \text{Volume}=(\text{ 2.63}\cdot2.19\cdot1.96)(cm^3) \\ \text{Volume}=11.289012(cm^3) \end{gathered}$

Step 2

now,

Let

$\begin{gathered} \text{Volume}=11.289012(cm^3) \\ \text{mass}=\text{ 45.3 gr} \end{gathered}$

replace in the formula

$\begin{gathered} \text{Density(d)}=\frac{mass(m)}{\text{volume(v)}} \\ d=\frac{45.3\text{ gr}}{11.289012(cm^3)} \\ d=4.012\frac{gr}{cm^3} \end{gathered}$

therefore, the answer is

$4.012\frac{gr}{cm^3}$

I hope this helps you

4 0

1 year ago

Whar determines the direction of ocean currents

Arturiano [62]

Answer:

The direction in which they rotate is determined by the hemisphere in which they are located. The main overall global wind patterns, the rotation of the Earth, and the shape of ocean basins are the three major factors that decide surface currents.

3 0

3 years ago

Read 2 more answers

How to find the value of sin(90+ theta) ?

koban [17]

Sin(90 + θ)

By addition angle formula for sine.

sin90cosθ + cos90sinθ

1*cosθ + 0*sinθ

cosθ + 0

= cosθ

sin(90 + θ) = cosθ

6 0

3 years ago