The sum of the initial and final velocity is divided by 2 to find the average.

Physics

1 answer:

4vir4ik [10]2 years ago

7 0

The average velocity is obtained by ( u + v)/2.

<h3>What is average velocity?</h3>

The term velocity refers to the rate of change of displacement with time. Velocity is a vector quantity thus it has both magnitude and direction. The velocity of a body could also be shown by the use of the vector notation.

To obtain the average velocity;

Let the initial velocity be u

Let the final velocity be v

It then follows that the average velocity = ( u + v)/2

Learn more about average velocity:brainly.com/question/862972

#SPJ1

You might be interested in

A 0.954-kg toy car is powered by three D cells (4.50 V total) connected directly to a small DC motor. The car has an effective e

Natasha2012 [34]

Answer:

0.06 A

Explanation:

We have given mass =0.954 kg

velocity =1.27 m/sec

efficiency =0.361 the output kinetic energy of the motor $KE=\frac{1}{2}mv^2=\frac{1}{2}\times 0.954\times 1.27^2=0.769\ J$

Efficiency =0.361

so input to the motor = output/efficiency

so input to the motor = $\frac{0.769}{0.361}=2.1301$

we know that $P=\frac{E}{T}$ where E is energy and T is time so $P=\frac{2.1301}{7.88}=0.2703W$

We know that power P=VI we have given V=4.5 VOLT

So current $I=\frac{P}{V}=\frac{0.2703}{4.5}=0.06 A$

3 0

3 years ago

Read 2 more answers

A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 c

valina [46]

Answer:

$q_1 =7.08*10^{-9}C.$

Explanation:

Gauss's Law says that the electric flux $\Phi_E$ through a closed surface is directly proportional to the charge $Q_{enc}$ inside it. More precisely,

$\Phi_E=\oint_S E\cdot dA = \dfrac{Q_{enc}}{\epsilon_0}.$

This means what is outside this closed surface $S$ does not contribute to the flux through it because field lines that go in must come out, <em>resulting a zero flux from an external charge. </em>

In our context, this means the charge $q_2$ which is outside the sphere will have zero flux through the surface; therefore, Gauss's law will only be concerned with charge $q_1$ which is inside the sphere; Hence,