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3 years ago

Which graph illustrates constant speed and velocity?

Physics

2 answers:

boyakko [2]3 years ago

8 0

The correct graph is D.

The graph A is a straight line sloping downwards and it shows that the speed of the body is decreasing at a constant rate. Therefore, this s a graph of a body that is under a constant deceleration.

The graph B is a straight line which slopes upwards. Hence the graph shows that the speed of the body increases at a constant rate. Therefore, this is a graph of a body that is accelerating at a constant rate.

The graph C is curved line, which curves upwards. The slope of the curve increases with time. This is therefore, a graph of a body which is under increasing acceleration.

The graph D, however is a straight line parallel to the time axis. The speed of the body has the same value at all times. Therefore, Graph D is the graph which shows the motion of a body with constant speed.

denpristay [2]3 years ago

5 0

Answer choice D is correct.

When you are thinking about a "constant" on a graph, it usually means that is is a straight line that is either horizontal or vertical. In this case the only graph that as both a constant speed and velocity is D which is why it is correct.

A has a negative constant acceleration

B has a positive acceleration

C is not constant, but is moving at a positive acceleration.

Hope this helps some!

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Which equation represents mass-energy equivalence? E = m2c E = mc2 E = (mc)2 E = mc

motikmotik

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$E = mc^2$

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4 0

3 years ago

Read 2 more answers

A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating

Rainbow [258]

Answer:

$41.4496148484\ m/s$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\sigma$ = Surface charge density = $5.9\times 10^{-8}\ C/m^2$

$\Delta x$ = 0.57-0.26

q = Charge = $6.9\times 10^{-9}\ C$

m = Mass of object = $8.3\times 10^{-9}\ kg$

Electric field due to a sheet is given by

$E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m$

Electric field is given by

$E=\dfrac{V}{d}$

Voltage is given by

$V=E\Delta x$

Kinetic energy is given by

$K=qV$

$\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s$

The initial speed of the object is $41.4496148484\ m/s$

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