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4 years ago

According to the second law of thermodynamics, what is true of entropy in all natural systems?

Physics

1 answer:

Stella [2.4K]4 years ago

8 0

Entropy is randomness.
It mostly applies to gas particles because they move around randomly and freely

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A bucket of water with mass 5 kg sits on the ground with a coefficient of static friction of 0.35. What is the maximum force of

allochka39001 [22]

Answer:

The force of static friction is 17.15 N

Explanation:

It is given that,

Mass of the bucket, m = 5 kg

The coefficient of static friction is, $\mu=0.35$

We need to find the maximum force of static friction. It is given by :

$F=\mu mg$

$F=0.35\times 5\ kg\times 9.8\ m/s^2$

F = 17.15 N

So, the force of static friction is 17.15 N. Hence, this is the required solution.

6 0

4 years ago

Read this quotation:

Brums [2.3K]

Answer:

the Princess in "St. George and the Dragon"

4 0

3 years ago

Read 2 more answers

a projectile whose masss is 10g is fired directly upward from ground level with an initial velocity of 1000m/s neglecting the ef

sammy [17]

Answer:

<em>The speed of the projectile when it impacts the ground is 1000 m/s</em>

Explanation:

<u>Vertical Launch</u>

When an object is launched vertically and upwards it starts to move at an initial speed vo, then the acceleration of gravity makes that speed to reduce until it reaches 0. The object has reached its maximum height. Then, it starts to move downwards in free fall, with initial speed zero and gradually increasing it until it reaches the ground level. We will demonstrate that the speed it has when impacts the ground is the same (and opposite) as the initial speed vo.

The speed when the object is moving upwards is given by

$v_f=v_o-g.t$

The time it takes to reach the maximum height is when vf=0, i.e.

$0=v_o-g.t$

solving for t

$\displaystyle t=\frac{v_o}{g}$

The maximum height reached is

$\displaystyle y=\frac{gt^2}{2}=\frac{v_o^2}{2g}$

Then, the object starts to fall. The object's height is given by

$\displaystyle y=\frac{v_o^2}{2g}-\frac{gt'^2}{2}$

where t' is the time the object has traveled downwards. The height will be 0 again when

$\displaystyle \frac{v_o^2}{2g}=\frac{gt'^2}{2}$

Solving for t'

$\displaystyle t'=\frac{v_o}{g}$

We can see the time it takes to reach the maximum height is the same it takes to return to ground level. Of course, the speed when it happens is

$v_f=g.t'=v_o$

Thus, the speed of the projectile when it impacts the ground is 1000 m/s

4 0

4 years ago

3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t

sattari [20]

Answer:

Work done is 12.3 J

Explanation:

We have,

Mass of puck, m = 0.35 kg

Force of friction acting on the puck when it slides is 0.15 N

Distance travelled by the puck is 82 m.

It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by

$W=Fd\\\\W=0.15\times 82\\\\W=12.3\ J$

The work done on the puck is 12.3 J.

5 0

4 years ago

Lindsay is planning a flight from St. Catherines to Hamilton, which lies due west of St. Catharines.

Schach [20]

Lindsay should fly the plane in the direction [W 12.5° S] to get Hamilton.

Using Sine rule to solve this question

Sine rule => SinA/a = SinB/b = SinC/c = constant

The magnitude of wind is 50 with an angle of 60 degrees.

The magnitude of plane is 200 and the angle at which it should fly is unknown and should be θ.

One side is 50 km/hr at an angle of 60 degrees.

sin 60°/200 = sin θ / 50

50 × sin 60° = 200 × sin θ

√3/2 = 4 × sin θ

√3/8 = sin θ

sin θ = 0.2165

θ = sin⁻¹(0.2165)

θ = 12.5°

So Lindsay have to fly the plane in the direction of [W 12.5° S].

Learn more about Sine Rule here:

brainly.com/question/27174058

#SPJ10

3 0

2 years ago