All categories

2 years ago

A position versus time graph is shown:

Physics

1 answer:

Stella [2.4K]2 years ago

8 0

Answer: Answer: The object moves forward at 5 m/s, stops, and then changes velocity.

Explanation:

With the information given in the question we can graph the points (image attached).

As we can observe, in the first segment of the graph the velocity is increasing linearly (at a constant rate) and is 5 m/s, then in the second segment we can see the position of the object remains the same from second 2 to second 4, which means the object is stopped.

Finally, in the third and last segment, we can observe a change in velocity (at a negative constant rate, because is decreasing), which is decreasing until the object stops.

Explanation:

You might be interested in

An object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m

Svetllana [295]

The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

<h3>Work done by the force experienced by the object</h3>

The magnitude of the work done by force experience by the object is calculated as follows;

W = f.d

where;

F is the applied force (2xyi + 3yj), where x and y are in meters
d is the displacement of the object = (a, b)

The work done by the force is determined from the dot product of the force and the displacement of the object.

F = (2xyi + 3yj).(a + b)

W = (2abi + 3bj).(ai + bj)

W = (2a²b + 3b²)J

Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

The complete question is below:

The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.

How much work does the force do?

Learn more about work done here: brainly.com/question/8119756

5 0

2 years ago

what will be the focal lenght of a combined lens made by contact of two lenses of power +3D and -2D.

4vir4ik [10]

P_1=+3D
P_2=-2D

$\\ \bull\tt\dashrightarrow P=P_1+P_2$

$\\ \bull\tt\dashrightarrow P=+3D-2D=+1D$

Now

$\\ \bull\tt\dashrightarrow f=\dfrac{1}{P}$

$\\ \bull\tt\dashrightarrow f=\dfrac{1}{1}$

$\\ \bull\tt\dashrightarrow f=1m$

3 0

2 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$