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3 years ago

Describe the importance of conservative forces to conservation of energy.

Physics

1 answer:

qwelly [4]3 years ago

6 0

Not sure but just coming to say good luck and take your time

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Answer ?

morpeh [17]

Answer:

2.75 m/s^2

Explanation:

The airplane's acceleration on the runway was 2.75 m/s^2

We can find the acceleration by using the equation: a = (v-u)/t

where a is acceleration, v is final velocity, u is initial velocity, and t is time.

In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1

a = 2.75 m/s^2

5 0

2 years ago

A girl is standing 150m in front of a tall building, fires a shot with starting pistol. A boy standing 350m behind her, hears tw

Natasha2012 [34]

Answer:

300 m/s

Explanation:

The difference in time between the two bangs is 1 s.

Thus;

t2 - t1 = 1

We know that distance/time = speed.

Thus;

d2/v - d1/v = 1

Multiply through by v to get;

d2 - d1 = v

Where v is speed of sound in air.

d1 = 350 m

d2 = (150 × 2) + 350 = 650 m

Thus;

v = d2 - d1 = 650 - 350 = 300 m/s

8 0

3 years ago

Which statement below could be used as evidence to support the claim that using nuclear energy to produce electricity is better

Alexxx [7]

Answer:

You didn't give the information needed for the answer bud

Explanation:

8 0

3 years ago

The difference between the speed of sound n air at 0°C and the speed of sound in air at 20°c is that... A. Cooler air molecules

sergejj [24]

Heat, like sound, is kinetic energy. Molecules at higher temperatures heave more energy, thus they can vibrate faster. Since the molecules vibrate faster, sound waves can travel more quickly.

So the answer is A.

7 0

3 years ago

Read 2 more answers

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago