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RoseWind [281]
3 years ago
12

State any 3 properties of an ideal gas as assumed by the kinetic theory.​

Physics
1 answer:
Rudik [331]3 years ago
5 0

Answer:

The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat.

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An object in motion will have a speed which is a scaler , or ( blank ) which is a vector .
mylen [45]

Answer:

Velocity

Explanation:

Objects in motion usually have a speed which is scalar or velocity which is a vector.

A scalar quantity is one with magnitude but has no directional attribute.

A vector quantity is one with both magnitude and directional attribute.

Speed is a scalar quantity that describes the magnitude of motion a body accrues.

Velocity is a vector quantity that describes both the magnitude of motion and the direction of motion in a body.

3 0
3 years ago
A drone is flying horizontally when it runs out of power and begins to free fall from 16 m. No drag. If it lands 40 m away (in t
marysya [2.9K]

Answer:

the horizontal velocity while it was falling is 22.1 m/s.

Explanation:

Given;

height of fall, h = 16 m

horizontal distance, x = 40 m

The time to travel 16 m is calculated as;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 16}{9.8} } \\\\t = 1.81 \ s

The horizontal velocity is calculated as;

v_x = \frac{X}{t} \\\\v_x = \frac{40}{1.81} \\\\v_x = 22.1 \ m/s

Therefore, the horizontal velocity while it was falling is 22.1 m/s.

7 0
2 years ago
How do you calculate the net force, i need a full explanation PLEASE
Lina20 [59]

Answer:

Once you have drawn the free-body diagram, you can use vector addition to find the net force acting on the object. We will consider three cases as we explore this idea:

Case 1: All forces lie on the same line.

If all of the forces lie on the same line (pointing left and right only, or up and down only, for example), determining the net force is as straightforward as adding the magnitudes of the forces in the positive direction, and subtracting off the magnitudes of the forces in the negative direction. (If two forces are equal and opposite, as is the case with the book resting on the table, the net force = 0)

Example: Consider a 1-kg ball falling due to gravity, experiencing an air resistance force of 5 N. There is a downward force on it due to gravity of 1 kg × 9.8 m/s2 = 9.8 N, and an upward force of 5 N. If we use the convention that up is positive, then the net force is 5 N - 9.8 N = -4.8 N, indicating a net force of 4.8 N in the downward direction.

Case 2: All forces lie on perpendicular axes and add to 0 along one axis.

In this case, due to forces adding to 0 in one direction, we only need to focus on the perpendicular direction when determining the net force. (Though knowledge that the forces in the first direction add to 0 can sometimes give us information about the forces in the perpendicular direction, such as when determining frictional forces in terms of the normal force magnitude.)

Example: A 0.25-kg toy car is pushed across the floor with a 3-N force acting to the right. A 2-N force of friction acts to oppose this motion. Note that gravity also acts downward on this car with a force of 0.25 kg × 9.8 m/s2= 2.45 N, and a normal force acts upward, also with 2.45 N. (How do we know this? Because there is no change in motion in the vertical direction as the car is pushed across the floor, hence the net force in the vertical direction must be 0.) This makes everything simplify to the one-dimensional case because the only forces that don’t cancel out are all along one direction. The net force on the car is then 3 N - 2 N = 1 N to the right.

Case 3: All forces are not confined to a line and do not lie on perpendicular axes.

If we know what direction the acceleration will be in, we will choose a coordinate system where that direction lies on the positive x-axis or the positive y-axis. From there, we break each force vector into x- and y-components. Since motion in one direction is constant, the sum of the forces in that direction must be 0. The forces in the other direction are then the only contributors to the net force and this case has reduced to Case 2.

If we do not know what direction the acceleration will be in, we can choose any Cartesian coordinate system, though it is usually most convenient to choose one in which one or more of the forces lie on an axis. Break each force vector into x- and y-components. Determine the net force in the x direction and the net force in the y direction separately. The result gives the x- and y-coordinates of the net force.

Example: A 0.25-kg car rolls without friction down a 30-degree incline due to gravity.

We will use a coordinate system aligned with the ramp as shown. The free-body diagram consists of gravity acting straight down and the normal force acting perpendicular to the surface.

We must break the gravitational force in to x- and y-components, which gives:

F_{gx} = F_g\sin(\theta)\\ F_{gy} = F_g\cos(\theta)F

gx

​

=F

g

​

sin(θ)

F

gy

​

=F

g

​

cos(θ)

Since motion in the y direction is constant, we know that the net force in the y direction must be 0:

F_N - F_{gy} = 0F

N

​

−F

gy

​

=0

(Note: This equation allows us to determine the magnitude of the normal force.)

In the x direction, the only force is Fgx, hence:

F_{net} = F_{gx} = F_g\sin(\theta) = mg\sin(\theta) = 0.25\times9.8\times\sin(30) = 1.23 \text{ N}F

net

​

=F

gx

​

=F

g

​

sin(θ)=mgsin(θ)=0.25×9.8×sin(30)=1.23 N

7 0
3 years ago
Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di
Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, u=20m/s

And the mass of the car is, m=1000 kg

The total distance covered by the car before stop, s=45m

And the final speed of the car is, u=0m/s

Now initial kinetic energy is,

KE_{i}=\frac{1}{2}mu^{2}

Substitute the value of u and m in the above equation, we get

KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J

Now final kinetic energy is,

KE_{f}=\frac{1}{2}mv^{2}

Substitute the value of v and m in the above equation, we get

KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN

Here, the force is negative because the force and acceleration in the opposite direction.

6 0
2 years ago
Which of the following plays a major role in creating deep ocean currents
iragen [17]

What are the options?

5 0
3 years ago
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