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3 years ago

Why can gases and liquids both transmit heat by convection?

Physics

2 answers:

defon3 years ago

6 0

Answer:

A. Their particles are free to move away from each other.

Explanation:

DochEvi [55]3 years ago

4 0

Answer:

Gases and liquids both transmit heat by convection because their particles are free to move away from each other (option A).

Explanation:

Convection is a process in which heat is transferred between two liquid substances, between two gaseous substances or a gas and a liquid when they are at different temperatures. The heat flux is produced from the body with the highest temperature to the substance with the lowest temperature.

The transmission of heat by convection is based on the actual movement of the molecules of a substance. Both liquids and gases are considered fluids where their particles are free to move away and move from each other. In solids particles vibrate around fixed positions and have a fixed volume .

Gases and liquids both transmit heat by convection because their particles are free to move away from each other (option A).

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Which statement best explains why radon and krypton do not bond easily with other elements?

mote1985 [20]

Answer.B

Explanation:

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3 years ago

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If a car accelerates from rest at a constant 4 m/s

Readme [11.4K]

Answer:

The time it will take for the car to reach a velocity of 28 m/s is 7 seconds

Explanation:

The parameters of the car are;

The acceleration of the car, a = 4 m/s²

The final velocity of the car, v = 28 m/s

The initial velocity of the car, u = 0 m/s (The car starts from rest)

The kinematic equation that can be used for finding (the time) how long it will take for the car to reach a velocity of 28 m/s is given as follows;

v = u + a·t

Where;

v = The final velocity of the car, v = 28 m/s

u = The initial velocity of the car = 0 m/s

a = The acceleration of the car = 4 m/s²

t = =The time it will take for the car to reach a velocity of 28 m/s

Therefore, we get;

t = (v - u)/a

t = (28 m/s - 0 m/s)/(4 m/s²) = 7 s

The time it will take for the car to reach a velocity of 28 m/s, t = 7 seconds.

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3 years ago

A photograph of you and your friends at your 8th birthday party.is it secondary or primary source and why?

Reptile [31]

Answer:

primary source

Explanation:

the explanation is in the image above

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3 years ago

You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho

timurjin [86]

Answer:

$\omega'=19.419\ rev.s^{-1}$

Explanation:

Given:

angular speed of rotation of friction-less platform, $\omega=5.1\ rev.s^{-1}$

moment of inertia with extended weight, $I=9.9\ kg.m^2$

moment of inertia with contracted weight, $I'=2.6\ kg.m^2$

Now we use the law of conservation of angular momentum:

$I.\omega=I'.\omega'$

$9.9\times 5.1=2.6\times \omega'$

$\omega'=19.419\ rev.s^{-1}$

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.

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3 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

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