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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 <u>D</u>.

The graph <em>A</em> 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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A 22kg box slides down a 30.0 degree incline. If there is no friction, what is the acceleration of the box?

Rashid [163]

Answer:

Acceleration of the box is 4.9 m/s².

Explanation:

The free body diagram is shown below.

The weight of the body can be resolved into two mutually perpendicular components as shown. The force responsible for sliding motion of the box along the incline is due to $mg\sin \theta$ acting along the incline.

Therefore, as per Newton's second law of motion,

Net force = $mass\times acceleration$

Now, net force acting along the incline is only $mg\sin \theta$

Therefore,

$mg\sin \theta=ma\\a=g\sin \theta$

Now, $g=9.8\textrm{ and } \theta = 30$

Therefore, the acceleration of the box is given as:

$a=9.8\times \sin(30)=9.8\times 0.5=4.9\textrm{ }m/s^{2}$

Acceleration of the box is 4.9 m/s².

6 0

4 years ago

A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude

OverLord2011 [107]

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

We would calculate the magnitude by applying pythagorean theorem:

$\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}$

$\longrightarrow \sf{Magnitude = 33.12}$

$\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}$

The angle between two vectors is given by the formula:

$\sf{\longrightarrow \small \cos \emptyset = \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }$

In two dimensional, the x axis of vector form is:

$\small\sf{\longrightarrow (b1, b2) = (1, 0) }$

$\sf{\longrightarrow \small \cos \: \emptyset = \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }$

$\small\longrightarrow \sf{ \dfrac{14}{33.12} }$

$\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}$

$\small\longrightarrow \sf{\emptyset= 115^\circ}$

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

$\small\bm{The \: angle \: between \: the \: vector \: }$

$\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}$

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2 years ago

(BRAINLIEST)

UkoKoshka [18]

Answer:

Gravity

because it's factorised by mass of a body.

For other forces, they deal with charges of negligible mass and weights

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3 years ago

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A positive and a negative charge are released from rest in vacuum. They move toward each other. As they do: A positive and a neg

Pachacha [2.7K]

Answer:

A negative potential energy becomes more negative

Explanation:

Let the charges be - Q₁ and Q₂ . Let the distance between them be d .

Potential energy = k -Q₁x Q₂ / R

= - KQ₁ Q₂ / R

Now if the magnitude of R decreases , the magnitude of potential energy increases . So we see that the negative potential energy becomes more negative .

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Answer:

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