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3 years ago

Which situation is an example of transferring heat by means of convection?

Physics

2 answers:

Nataly_w [17]3 years ago

6 0

Answer: b. the weather patterns outside

Explanation:

There are three modes of heat transfer:

1. Conduction: This type of heat transfer happens when there is direct contact between the two object.

2. Convection: This type of heat transfer happens when there is a movement of fluid (liquid or gas).

3. Radiation: This type of heat transfer happens when there is direct transfer of energy through space.

a. The sun warming the earth is through radiation as there is no direct contact.

b. The weather patterns outside are created due to the convection currents which arises due to movement of hot fluid upwards due to less density and the cold fluid downwards due to more density.

c. You toasty-warm in some footy pajamas is an example of conduction as there is direct contact.

d. You accidentally grabbing the hot pan from the oven is an example of conduction as there is direct contact.

kondaur [170]3 years ago

5 0

<span>b. the weather patterns outside

There are three types of heat transfer or heat propagation; conduction, convection and radiation. Heat transfer is the process by which heat projects externally however, depending on the temperature and pressure. Also called the movement of heat from a low temperatured state which increases as heat progresses.
Conduction is the heat transfer by contact, immediate contact.
Convection is the transfer of heat through air and water.
Radiation is the transfer of heat regardless of the presence of atoms or particles.

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A cat is shot horizontally out of a cannon that is on top of a 125 meter tall cliff. If the cat lands 286 m away from the base o

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The initial velocity of the cat is 56.6 m/s

Explanation:

The motion of the cat is a projectile motion, consisting of:

- a uniform horizontal motion with constant velocity

- a vertical accelerated motion with constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time it takes for the cat to reach the ground. We use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where we have (choosing downard as positive direction)

s = 125 m is the vertical displacement of the cat

u = 0 is the initial vertical velocity

t is the time taken to reach the ground

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

Solving for t, we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(125)}{9.8}}=5.05 s$

Now we can analzye the horizontal motion. Since the motion is uniform and the horizontal velocity is constant, it is given by

$v_x =\frac{d}{t}$

where

d = 286 m is the horizontal distance travelled

t = 5.05 s is the time taken

Solving the equation,

$v_x = \frac{286}{5.05}=56.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

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3 years ago

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

n200080 [17]

a) $8.93\cdot 10^{-30} N$ , downward

b) $1.76\cdot 10^{-17} N$ , upward

c) $1.20\cdot 10^{-17} N$ , downward

Explanation:

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

$F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N$

And the direction is downward, towards the Earth's centre.

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$E=110 N/C$ is the strength of the electric field

Substituting,

$F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N$

And since the electron has negative charge, the direction of the force is opposite to that of the electric field, so upward.

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

$F=qvB$

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$v=1.50\cdot 10^6 m/s$ is the velocity

$B=50.0\mu T = 50.0 \cdot 10^{-6} T$ is the strength of the magnetic field

Substituting,

$F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N$

And the direction can be determined using the right-hand rule:

- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

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3 years ago