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3 years ago

Over time, how do the matter and energy in a system naturally change?

Physics

2 answers:

natta225 [31]3 years ago

6 0

Answer:

A. Their entropy increases

Explanation:

The total entropy of a system either increases or remains constant in any process; it never decreases.

Marina86 [1]3 years ago

3 0

I’m going to say that it A but I can be wrong.

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Describe all the ways that newtons laws can apply in a car crash

Harman [31]

Newtons first law - Objects in the car at rest (The human) will remain at rest unless affected by an unbalanced force. Well the unbalanced force would be the crash and this would set the human in motion and they would ether fly out the car if not wearing a seat belt or if wearing one they would get bad whip lash

Newtons second law - With more mass requires more force, so since the human is pretty light or even if heavy in a big crash there will be so much more from it that this will send the human flying.

Newtons 3rd law - Objects A puts force onto objects b and object b excretes the same amount of force back onto object a, so in a crash the human would hit the car hard and the car would excrete the same amount of force back on the human which would really damage him/her

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3 years ago

New stars form from ?

allsm [11]

Stars form from an accumulation of gas and dust, which collapses due to gravity and starts to form stars. The process of star formation takes around a million years from the time the initial gas cloud starts to collapse until the star is created and shines like the sun

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3 years ago

A student says that a speed of 50 m/s is faster than a speed of 140 km/h because the number is bigger. What would you say to the

denis23 [38]

Yes you are right but sometimes just because a number is bigger doesnt always make it so.

5 0

3 years ago

1. A 20m steel wire is stretched to 20.0m by

Eduardwww [97]

Answer:

Force to stretch the wire is 250 N

Explanation:

As we know that modulus of elasticity will remain the same for the wire if the applied stretch to the wire is within elastic limit

So we will have

$\frac{F L}{\Delta L A} = constant$

now we have

$\frac{F_1 L}{\Delta L_1 A} = \frac{F_2 L}{\Delta L_2 A}$

so we can write it as

$F_2 = \frac{F_1 L_2}{L_1}$

$F_2 = \frac{50 (0.05)}{0.01}$

$F_2 = 250 N$

7 0

3 years ago

A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0

4 years ago