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3 years ago

Difference between uniform motion and non uniform motion

Physics

2 answers:

photoshop1234 [79]3 years ago

5 0

Answer:

Uniform motion is a type of motion that is characterized as the motion of an object wherein the object moves in a straight line and its velocity remains unchanged along that line, regardless of the duration of time, as it occupies equal distances at equal time interval and Non-uniform motion is described as the motion of an object wherein the object moves at different speeds and does not cover the same distance at equal time intervals, regardless of the duration of the time interval.

oee [108]3 years ago

4 0

Answer:

hope it's help you

Explanation:

In Uniform motion, the movement of a body is along the straight line with constant speed. In non uniform motion, the movement of a body is along the straight line with variable speed. In uniform motion, the body covers equal distance in an equal interval of time.

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Which could describe the motion of an object?

Volgvan

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3 0

3 years ago

Read 2 more answers

A space station with a radius of 120 m rotates once every 70 s to create artificial gravity. If the astronaut has an earth weigh

Alina [70]

Answer:

Artificial weight = 70.27 N = 15.80 lbs

Explanation:

The earth weight of the astronaut = 160 lbs = 711.72 N

The weight on earth = m × g(earth)

g(earth) = 9.8 m/s²

711.72 = m × 9.8

m = (711.72/9.8)

m = 72.62 kg

But at the space station, the space station rotates once every 70 s to create an artificial radial acceleration that creates a radial gravity pulling the objects on the space station towards the centre of that space station.

radial acceleration = α = (v²/r)

v = rw,

α = (rw)²/r

α = rw²

r = radius of rotation = 120 m

w = angular velocity = (2π/70) (it completes 1 rotation, 2π radians, in 70 s)

w = 0.0898 rad/s

α = 120 × (0.0898²)

α = 0.968 m/s²

Artificial weight = (mass of astronaut) × (Radial acceleration) = 72.62 × 0.968

Artificial weight = 70.27 N = 15.80 lbs

Hope this Helps!!!

5 0

4 years ago

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire

Deffense [45]

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$ , that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning: