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4 years ago

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Larry designed an experiment to show how heat can be transferred from one place to another. The steps of the experiment are show

n below. Light a candle. Place palm of the hand about 6 inches above the flame. Feel the heat from the burning candle on the palm of the hand. What does Larry's experiment most likely demonstrate?

1 answer:

marysya [2.9K]4 years ago

8 0

What Larry's experiment most likely demonstrate is <span>heat transfer by convection

</span>Heat transfer by convection occurs when particles of heated liquid or gases travel<span> away from the source, carrying thermal energy along. </span>Convection<span> above the flame of the lit candle occurs because the hot air from the candle expands, becomes less dense, and rises towards the palm.</span>

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choose a substance you're familiar with. what are its physical and chemical properties? How would you measure its density? What

Artist 52 [7]

You can look at magnesium, it can react with oxygen to form oxides. (chemical) it is malleable and a solid at room temperature. (physical)

to measure its density, the mass and volume can be worked out and from this density too. look up the equation, it is quite easy :)

physical changes -- it can be melted, and oxidized <span />

7 0

4 years ago

Read 2 more answers

Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 57.5 cm from the slit.

boyakko [2]

Answer:

$71.0 \mu m$

Explanation:

The formula for the single-slit diffraction is

$y=\frac{n\lambda D}{d}$

where

y is the distance of the n-minimum from the centre of the diffraction pattern

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light

In this problem,

$\lambda=648.0 nm=6.48\cdot 10^{-7}m$

$D=57.5 cm=0.575 m$

$y=1.05 cm=0.0105 m$ , with n=2 (this is the distance of the 2nd-order minimum from the central maximum)

Solving the formula for d, we find:

$d=\frac{n\lambda D}{y}=\frac{2(6.48\cdot 10^{-7} m)(0.575 m)}{0.0105 m}=7.10\cdot 10^{-5} m= 71.0 \mu m$

6 0

3 years ago

A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, t

Brilliant_brown [7]

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

$y=\frac{1}{2}at^2$

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

$y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m$

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s$

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}$

6 0

3 years ago

We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in

leva [86]

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

$B=\mu_0\, \frac{N}{L} I$

Then, if we assign the subindex "1" to the quantities that define the magnetic field ( $B_1$ ) inside solenoid 1, we have:

$B_1=\mu_0\, \frac{N_1}{L_1} I_1$

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then $L_2=L_1/2$

2) twice as many turns as solenoid 1. Then $N_2=2\,N_1$

3) three times the current of solenoid 1. Then $I_2=3\,I_1$

we obtain:

$B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1$

5 0

4 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

4 years ago