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8_murik_8 [283]
3 years ago
5

Heat always flows from a _________ object to a __________ one.

Physics
2 answers:
PolarNik [594]3 years ago
8 0

From a hot object to a cold object. Ill give an example as to why:

Say you place an ice cube in a glass of warm water, we know that in colder substances, the average kinetic energy per particle (which is the classical definition for temperature is lower). In hotter substances the average kinetic energy per particle is larger, as it is hotter. So now the ice is placed in the warm water. The fast moving particles within the water, that have higher kinetic energies will bounce off the slower moving particles in the ice with lower average kinetic energies. This causes an exchange in energy as the substance with more average kinetic energy per particle (which is temperature) will transfer energy to the slower moving particles located in the ice. This will eventually cause the water and ice to reach an equilibrium temperature as a result of the exchange, with the energy taken to melt the ice included.

satela [25.4K]3 years ago
6 0

Heat always flows from a B. hotter object to a colder one.

Heat transfers always from hotter to colder.

~

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Which of the following classifications of star temperature is coolest?
Oliga [24]
The hottest would be the O type and the coolest is M
3 0
3 years ago
Read 2 more answers
How does the gravitational force between two objects change if the distance between the objects doubles?
Ymorist [56]

if the distance between the objects is doubled the force is reduced by a factor of 4

<h3>Whats is gravitational force?</h3>

Gravitational force is the force of attraction between objects in the universe.

f = G * m1 * m2 / r^2

f = gravitational force

G = gravitational constant

m1 = mass of object 1

m2 = mass of object 2

r = distance between the objects

From the formular, the gravitational force and the distance is an inverse relationship so increasing the distance by a factor results to reduction of the force by the square of the factor. hence doubling the distance which is distance mutiplied by 2 will lead to reduction of the force by 2^2 = 4

Therefore: The force decreases by a factor of 4.

hope it helps

4 0
2 years ago
Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5
LekaFEV [45]

Answer:

  • |\vec{F}_3| = 102.92 \ N
  • \theta = 57 \° 24 ' 48''

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

\vec{a} = \vec{0}.

As the net force equals acceleration multiplied by mass , this must mean:

\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}.

So, the sum of the three forces must be zero:

\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0},

this implies:

\vec{F}_3  = - \vec{F}_1 - \vec{F}_2.

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector \hat{i} pointing east,  with the y axis unit vector \hat{j} pointing south, so the positive angle is south of east. For this, we got for the first force:

\vec{F}_1 = 83.7 \ N \ (-\hat{j}),

as is pointing north, and for the second force:

\vec{F}_2 = 59.9 \ N \ (-\hat{i}),

as is pointing west.

Now, our third force will be:

\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})

\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}

\vec{F}_3  =  (59.9 \ N , 83.7 \ N )

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}

|\vec{F}_3| = 102.92 \ N

this is the magnitude.

To find the direction, we can use:

\theta = arctan(\frac{F_{3_y}}{F_{3_x}})

\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })

\theta = 57 \° 24 ' 48''

and this is the angle south of east.

7 0
3 years ago
The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.
ivolga24 [154]

The De Broglie's wavelength of a particle is given by:

\lambda=\frac{h}{p}

where

h=6.6 \cdot 10^{-34} Js is the Planck constant

p is the momentum of the particle


In this problem, the momentum of the electron is equal to the product between its mass and its speed:

p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m


So, the answer is True.

7 0
3 years ago
An object completes one and half revolution of a circle of radius R calculate the displacement and distance
Papessa [141]

Answer:

Distance is path length covered by particle. When particle moves along half circle, it covers half the circumference therefore distance covered is (2×pi×r)/2 = pi× r. ... Hence displacement is equal to diameter or 2 times the radius of circle.

8 0
3 years ago
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