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3 years ago

A positive or negative outcome of a decision or action can be called a

1 answer:

7 0

The answer to this would be :
<span>consequence
</span>hope that this helps you!
Source: I had done a research about this. =)

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Arocket with an initial velocity of 20 m/s another engine that gives it an acceleration of 4 m/s ^ 2 over 10 secondsHow far did

ExtremeBDS [4]

Answer:

50m

Explanation:

Given parameters:

Initial velocity = 20m/s

Acceleration = 4m/s²

Time = 10s

Unknown:

Distance traveled by the rocket = ?

Solution:

To solve this problem use the expression below;

v² = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

final velocity = 0

Insert the parameters and solve;

0² = 20² + 2 x 4 x s

-400 = 8s

s = 50m

Disregard the negative sign because distance cannot be negative.

3 0

2 years ago

An object initially at rest experiences an acceleration

lidiya [134]

Answer:

1.4m/s

Explanation:

Average velocity is the total distance covered divided by the total time taken.

Average velocity = $\frac{total distance }{time }$

Total time taken = 5s + 6s = 11s

The first distance covered = velocity x time = 1.4 x 5 = 7m

second distance covered = velocity x time = 1.4 x 6 = 8.4m

So;

Average velocity = $\frac{7 + 8.4}{11}$ = 1.4m/s

5 0

3 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0

3 years ago

If you decrease the distance between successive crests of a wave, this changes

AlexFokin [52]

<span>The _______ is the the distance between two crests or two troughs on a transverse wave. It is also the distance between compressions or the distance between rarefactions on a longitudinal wave.</span>

3 0

2 years ago