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3 years ago

Please solve it ,,,,,,,,,,................................

Physics

1 answer:

fredd [130]3 years ago

6 0

Answer:

the distance covered by the body in 5th second =u+a(n- 1/2)=7+4(5- 1/2)

=7+4×9/2

=7+18

=25m

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An electromagnetic wave is transporting energy in the negative y direction. At one point and one instant the magnetic field is i

natali 33 [55]

Answer:. Option c

Explanation: the speed of an electromagnetic wave is simply the vector product of the magnetic field and the electric field.

The direction of the velocity is the direction of the electromagnetic wave.

The wave is already moving towards the negative y axis (-j) and the magnetic field is already pointing towards the positive x axis (i)

From cross product of unit vectors

i × j = k

i × k = - j

With the second identity, we can see that the electric field will be pointing towards the positive of the x axis (k).

Option c is validated

4 0

2 years ago

Which equation represents a neutralization reaction?

elixir [45]

B acid + base → water + salt

Explanation:

The best equation that represents a neutralization reaction is:

acid + base → water + salt

In a neutralization reaction, an acid reacts with a base to produce salt and water only.

Here, hydrogen combines with hydroxide to form water.

Titration is usually used to carry out this reaction in the laboratory.

An indicator is added to the base and at end the color of the reaction changes.

At this point the acids have completely been neutralized with the base.

Learn more:

Neutralization brainly.com/question/4455839

#learnwithBrainly

5 0

3 years ago

Two point charges of -7uC and 4uC are a distance of 20

aivan3 [116]

Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy $\mathrm{EPE}$ .

$\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r}$ ,

where

The coulomb's constant $k = 8.99\times 10^{9}\; \rm N\cdot m^{2} \cdot C^{-2}$ ,
$q_1$ and $q_2$ are the sizes of the two charges, and
$r$ is the separation of (the center of) the two charges.

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

$q_1 = -7\rm \;\mu C = -7\times 10^{-6}\; C$ ;
$q_2 = 4\rm \;\mu C = 4\times 10^{-6}\; C$ ;

Initial separation: $\rm 20\; cm = 0.20\; cm$ ;
Final separation: $\rm 90\; cm = 0.90\; cm$ .

Apply Coulomb's law:

Initial potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}$ .

Final potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}$ .

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

$\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}$ .