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3 years ago

Which equation represents the law of conservation of energy in a closed system?

Physics

2 answers:

sladkih [1.3K]3 years ago

8 0

Answer:

A: KEi + PEi = KEf + PEf

Explanation:

on edge

brainliest will be appreciated

Irina-Kira [14]3 years ago

7 0

Answer:

KE + PE = KE + PE

Explanation:

In a closed system, the mechanical energy of the system is constant.

Mechanical energy is given by the sum of kinetic energy and potential energy; mathematically:

U = KE + PE

where

KE is the kinetic energy

PE is the potential energy

This means that if we consider two situations, one at the beginning and one at the end, the value of U will not change if the system is closed; this means that the sum KE + PE will remain the same, so we can write:

KE + PE = KE + PE

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A concert loudspeaker suspended high off the ground emits 28.0 W of sound power. A small microphone with a 0.700 cm2 area is 55.

grandymaker [24]

Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically, I = $\frac{P}{A}$

As it is given that power is 28.0 W and area is $7 \times 10^{-5} m^{2}$ .

Therefore, sound intensity will be calculated as follows.

I = $\frac{P}{A}$

= $\frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}$

= $0.318 \times 10^{5} W/m^{2}$

or, = $3.18 \times 10^{4} W/m^{2}$

Thus, we can conclude that sound intensity at the position of the microphone is $3.18 \times 10^{4} W/m^{2}$ .

7 0

2 years ago

Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t

elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0

h = 1/2 gt²

$t = \sqrt{\frac{2h}{g} }$

Horizontal distance covered = t x Vx

= $\sqrt{\frac{2h}{g} } \times V_x$

From this expression also

Horizontal distance covered is proportional to $\sqrt{h}$ .

7 0

3 years ago

4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive

zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

B1 at 20km/h

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0

3 years ago

An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang

kirza4 [7]

Answer:

$a=500m/s^2$

Explanation:

We need only to apply the definition of acceleration, which is:

$a=\frac{v_f-v_i}{t_f-t_i}$

In our case the final velocity is $v_f=10m/s$ , the initial velocity is $v_i=0m/s$ since it departs from rest, the final time is $t_f=0.02s$ and the initial time we are considering is $t_i=0s$