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2 years ago

Which of the following is an example of convection

Physics

2 answers:

Fantom [35]2 years ago

6 0

Answer:

Hot air rising and cooler air falling

Explanation:

hot air rising and cooler air falling

These are called "convection currents", the currents being air currents which carry heat because they are hot air - rises - and cold air - sinks. Also may explain coastal land and sea breezes. As "thermals" convection currents may help aeroplane gliders and birds to fly by "hitching a ride on the nearest convection current" as it were

natta225 [31]2 years ago

4 0

Answer:

Boiling water - The heat passes from the burner into the pot, heating the water at the bottom. Then, this hot water rises and cooler water moves down to replace it, causing a circular motion.

Radiator - Puts warm air out at the top and draws in cooler air at the bottom.

Steaming cup of hot tea - The steam is showing heat being transfered into the air.

Ice melting - Heat moves to the ice from the air. This causes the melting from a solid to liquid.

Hot air balloon - A heater inside the balloon heats the air and so the air moves upward. This causes the balloon to rise because the hot air gets trapped inside. When the pilot want to descend, he releases some of the hot air and cool air takes it place, causing the balloon to lower.

Frozen material thawing - Frozen food thaws more quickly under cold running water that if it is placed in water. The action of the running water transfers heat into the food faster.

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Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago

A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

$\boxed {I = \Sigma F \times t}$

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

$Ek = 70\% \ Ek_o$

$\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )$

$v^2 = 70 \% \ (v_o)^2$

$v = - v_o \sqrt{70 \%}$ → <em>negative sign due to ball rebounds</em>

$v = - v_o \sqrt{0.7} \texttt{ m/s}$

$\texttt{ }$

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

$\Delta p_{stone} = - \Delta p_{ball}$

$\Delta p_{stone} = - [ mv - mv_o ]$

$\Delta p_{stone} = m[ v_o - v ]$

$\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]$

$\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}$

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Average Speed of Plane : brainly.com/question/12826372
Impulse : brainly.com/question/12855855
Gravity : brainly.com/question/1724648

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0

2 years ago

Frequency is deoted as hertz; hertz is a measurement of the _________ _____ __________ that a wave is occurring.

nlexa [21]

Hertz is a measurement of the frequency that a wave is occurring.

4 0

2 years ago

Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc

lubasha [3.4K]

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m= mass of the substance (unity: grams or g), and

ΔT= difference of temperature (unity: Celsius degrees or °C); and

m= 95g and ΔT= 16°C:

Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal </u>

8 0

2 years ago

You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number

mamaluj [8]

The mass of an electron, me, 9.11 × 10⁻³¹ kilograms shows there are 3 important numbers in them, namely numbers 9, 1, and 1

<h3><em>Further explanation</em></h3>

Significant numbers are obtained from measurement results of exact numbers and the last number estimated

This is called significant numbers

the scientific notation can be described as:

a, ... x 10ⁿ

a, ... called a significant number

10ⁿ is called a big order

Rules for significant numbers in general:

1. All non-zero numbers are significant numbers

2. a zero which is located between two non-zero numbers including a significant number

3. all zeros are located in the final row written behind the decimal point of the include significant number

4. zero decimal point is the not significant number

From the numbers known in the question amounting to 9.11 × 10⁻³¹ kilograms shows there are 3 important numbers in them,(symbol a, before a big order ,called a significant number) namely numbers 9, 1, and 1

<h3><em>Learn more</em></h3>

significant figures

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significant figures in the following number: 5.67 x 106

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the number of significant figures is 0.025

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0.080 significant figures

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the fewest number of significant figures

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Keywords : significant number, the scientific notation, decimal point, a big order