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3 years ago

When velocity is graphed with respect to time, what does the area under

Physics

2 answers:

denpristay [2]3 years ago

6 0

The Answer

Is Displacement

Juliette [100K]3 years ago

4 0

Answer:

the answer is D

Explanation:

Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.

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¿cuales son los motivos para hacer una carta formal?<br>es para hoy porfavor necesito su ayuda

galina1969 [7]

La carta motivos es un documento escrito dentro del formato epistolar, por el cual se expresan las razones por la cual un postulante solicita algo, por ejemplo: continuar con sus estudios de posgrados en una universidad fuera de su ciudad natal o para dar una razón por la cual se encuentra calificado para un puesto de trabajo.

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8 0

2 years ago

Read 2 more answers

For the circuit shown in the figure(figure 1) find the current through each resistor. Express your answers using two significant

Angelina_Jolie [31]

The current flowing in each resistor of the circuit is 4 A.

<h3>Equivalent resistance of the series resistors</h3>

The equivalent resistance of the series circuit is calculated as follows;

6 Ω and 4 Ω are in series = 10 Ω

5 Ω and 10Ω are in series = 15 Ω

<h3>Effective resistance of the circuit</h3>

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \\\\R = \frac{R_1R_2}{R_1 + R_2} \\\\R = \frac{10 \times 15}{10 + 15} \\\\R = 6 \ ohms$

<h3>Current flowing in the circuit</h3>

V = IR

I = V/R

I = 24/6

I = 4 A

Learn more about resistors in parallel here: brainly.com/question/15121871

8 0

2 years ago

What happens when sound waves move faster?

Lady bird [3.3K]

Hnjjjjj hejehtjriwje a wkirrhnemekwhr. Rthejb wine rnrie

6 0

2 years ago

A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 Î¼C on each plate. The plates have a

butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance $C=260\ pF$

Charge $q=0.155\ \mu\ C$

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

$V= \dfrac{Q}{C}$

Where, Q = charge

C = capacitance

Put the value into the formula

$V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}$

$V=596.2\ volts$

Hence,The potential difference between the plates is 596.2 volts.

7 0

3 years ago

Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per

Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90 $^{\circ}$ to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

7 0

3 years ago