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3 years ago

Does this curved graph show a function explain how you know

Physics

1 answer:

pochemuha3 years ago

8 0

Answer:

No; the graph fails the vertical line test.

The vertical line test is a tool used to determine if we have a function. If we can draw a single straight vertical line through more than one point on the red curve, then the graph is said to have failed the vertical line test. Consequently, this leads to the relation not being a function.

For this circle graph, we can draw a vertical line through more than one point, which is why we don't have a function here.

Put another way, there are inputs (x) that produce more than one output (y), so that's why we don't have a function.

-BBBM

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Sonrisa owns a 300 W television. If the total energy usage for February is 32.4 kWh, how many hours per week does Sonrisa watch

Zielflug [23.3K]

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3 years ago

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Which object represents a negatively charged particle? which object represents a positively charged molecule? which object repre

kari74 [83]

The answers to your questions are as written below:

The objects that represents a negatively charged particle is : Image B
The object that represents a positively charged molecule is : Image A
The object that represents an uncharged molecule is : Image C
The object the will not move when in an electric fied is : Image C

<h3>Different types of charges molecules</h3>

A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.

A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.

Hence we can concude that the answers to your questions are as listed above.

Learn more about electric charges :brainly.com/question/857179

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8 0

2 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

2 years ago

Explain the relationship between mass and gravity

Pepsi [2]

Answer:

Since the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force. So as the mass of either object increases, the force of gravitational attraction between them also increases.

Explanation:

i did some research and this is what I got. hope it helps.

8 0

2 years ago

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In a system with only a single force acting upon a body, what is the relationship between the change in kinetic energy and the w

Makovka662 [10]

Answer: W.D = 1/2mv^2

Explanation:

If an external force or a single force is acting on a body. Just like the first law of thermodynamics, the force acting on the body will cause work done on the system.

Work done = force × distance

And the work done on the body will cause the molecules of the body to experience motion and thereby producing kinetic energy.

The work done will be converted to kinetic energy.

W.D = 1/2mv^2

6 0

3 years ago