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When an object with a negative charge is moved from point A to point B through an external electrical field, it gains electrical

Sergio039 [100]

Explanation:

As per the problem,

$\Delta U = (V_{B} - V_{A})(-q) > 0$

When q > 0 then -q is a negative charge . Since, change in potential energy ( $\Delta U$ ) increases.

or, $(V_{A} - V_{B})q$ > 0

or, $V_{A} > V_{B}$

Therefore, both positive and negative charge will move from $V_{A}$ to $V_{B}$ and as $V_{B} < V_{A}$ so both of them move through a negative potential difference.

Thus, we can conclude that the true statements are as follows.

The positively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).

The negatively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).

4 0

3 years ago

A rocket at rest with a mass of 942 kg is acted on by an average net force of 6,731 N upwards for 21 s. What is the final veloci

Galina-37 [17]

Answer:

Explanation:

Givens

m = 942

F = 6731

t = 21 seconds

vi = 0

vf = ?

Formula

F = m * (vf - vi ) / t

Solution

6731 = 942*(vf - 0)/21 Multiply both sides by 21

6731 * 21 = 942*vf

141351 = 942*vf Divide by 942

141351/942 = vf

vf = 151 m/s

6 0

3 years ago

A 1210 kg rollercoaster car is

ratelena [41]

Answer: 4.98 m/s

Explanation:

You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.

PEi = 0 in this case

KEi = ½mVi² = PEf+KEf = mghf + ½mVf²

½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²

½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26

Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s

3 0

3 years ago

Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface

salantis [7]

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

R = 2r

= $2 \times 0.1 m$

= 0.2 m

where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

U = $\frac{k \times Q \times Q}{R}$

= $\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}$

= 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, $\frac{U}{2}$ .