El empuje que sufre un cuerpo en un liquido es igual al pseo de volumen desalojado

KIM [24]

That's the impression you get from the sound, that you call 'high' or 'low', determined by the FREQUENCY of the wave.

3 0

3 years ago

Professor Whitney has a sample of lead and a sample of iron. The samples have equal mass. When Whitney heats the samples, the le

xenn [34]

The temperature increase of a substance is T=Q/m*c, where m is the mass, Q is the energy absorbed and c is the specific heat. So you can conclude that if the lead gets to a higher temperature, it must have a lower specific heat

8 0

4 years ago

*please refer to photo attached* The figure below shows a small, charged sphere, with a charge of q = +44.0 nC, that moves a dis

pishuonlain [190]

(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

(d) The potential difference between A and B is 56.7 V.

<h3>Electric force on the sphere</h3>

The electric force on the sphere is calculated as follows;

F = Eq

where;

E is electric field
q is the charge

F = 300 x (44 x 10⁻⁹)

F = 1.32 x 10⁻⁵ N

The direction of the force is towards the right.

<h3>Work done on the sphere</h3>

W = Fd

W = 1.32 x 10⁻⁵ N x 0.189 m

W = 2.5 x 10⁻⁶ J

<h3>Change of the electric potential energy </h3>

The change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.

<h3> Potential difference between A and B</h3>

VB − VA = Ed

VB − VA = 300 N/C x 0.189 m

VB − VA = 56.7 V

Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

The potential difference between A and B is 56.7 V.

Learn more about potential difference here: brainly.com/question/9060304

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6 0

2 years ago

Two equipotential surfaces surround a +1.70 x 10-8-C point charge. How far is the 120-V surface from the 54.0-V surface?

UNO [17]

Answer:

1.55 m

Explanation:

The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:

$V =\frac{k*q}{r}$

As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.

Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:

$r1 =\frac{k*q}{V1} = \frac{9e9 N*m2/C2*1.7e-8C}{120 V} = 1.28 m$