All categories

3 years ago

6. Draw a velocity-time graph for an object originally traveling at -3 m/s. The object

Physics

1 answer:

faltersainse [42]3 years ago

4 0

See the graph in attachment

Explanation:

In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.

In the graph, we see that the initial velocity at time t = 0 is

$v_0 = -3 m/s$

and it is negative, so below the x-axis.

Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).

After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

You might be interested in

A car travels 3 km north, then turns and travels 4 km west. What is the car's displacement?

Olin [163]

Explanation:

Displacement = Shortest path Travelled from starting point to final point..

Displacement formula = Final position - Initial position

= 7km - 3km

= 4km

^Displacement is 4km

If this answer helps you plz mark as brainlist..

Tq..

5 0

3 years ago

Read 2 more answers

Divergent faults form when tectonic plates _____.

Ronch [10]

move away from one another

(convergent is the opposite way)

5 0

3 years ago

A 500 lines per mm diffraction grating is illuminated by light of wavelength 580 nm . what is the maximum diffraction order seen

vampirchik [111]

The maximum diffraction order seen is 3.

<h3>What is the maximum diffraction order seen?</h3>

We know that the maximum angle of diffraction Q_m of the furthest bright fringe < Q = 90 degrees.

Here we need to compute the nth bright fringe for which is approximated to 90 degrees.

The angle of nth bright fringe is given by;

sin(Q_m) = n(λ)N

Approximating Q_m ≈ 90 degrees.

sin (90) = nλN

n = sin (90) / (λN)

n = 1 / ((580 x 10⁻⁶)500)

n = 3.5 orders

Since, we knew that Q_m < 90 degrees, we will choose n = 3 as the maximum number of orders.

Thus, the maximum diffraction order seen is 3.

Learn more about maximum diffraction here: brainly.com/question/14703089

#SPJ4

6 0

2 years ago

Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (

alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b) $E_1=\dfrac{\sigma}{\epsilon_0}$

c) $E_2=\dfrac{\sigma}{\epsilon_0}$

Explanation:

Let $\sigma$ be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

$E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}$

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

$E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0$

b)The Electric Field to the right of the sheets

$E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

c)The Electric Field to the left of the sheets

$E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

3 0

3 years ago

Action and Reaction are equal in magnitude and opposite in direction.Then,Why do they not balance each other?

Ivan

Answer:

Newton's third law of motion states that every action has an equal and opposite reaction. This indicates that forces always act in pairs. Reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel each other out.

4 0

2 years ago