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3 years ago

Which statements does the Second Law of Thermodynamics support?

1 answer:

4 0

According to the second law of thermodynamics,
the answer is
<span>4. The entropy of the universe is increasing. </span>

You might be interested in

A car accelerates from rest at a constant rate of 2m/s2 for 5s. what is the speed of a car at the end of that time?

Fiesta28 [93]

Good afternoon.

We have:

$\mathsf{V_0 = 0}\\ \mathsf{a = 2 \ m/s^2}\\ \mathsf{t = 5 \ s}$

The function of velocity:

$\mathsf{V = V_0+at}\\ \\ \mathsf{V = 0 + 2t}\\ \\ \mathsf{V = 2t}$

For t = 5 s:

$\mathsf{V = 2\cdot 5}\\ \\ \boxed{\mathsf{V = 10 \ m/s}}$

4 0

3 years ago

A house is lifted from its foundations onto a truck for relocation. The house is pulled upward by a net force of 2850 N. This fo

Gwar [14]

Answer:

m = 95000 kg

Explanation:

Given that,

Net force acting on the house, F = 2850 N

Initial speed, u = 0

Final speed, v = 15 cm/s = 0.15 m/s

We need to find the mass of the house. Let the mass be m. We know that the net force is given by :

F = ma

Where

a is the acceleration of the house.

So,

$F=m\dfrac{v-u}{t}\\\\m=\dfrac{Ft}{(v-u)}\\\\m=\dfrac{2850\times 5}{(0.15-0)}\\\\m=95000\ kg$

So, the mass of the house is equal to 95000 kg.

3 0

3 years ago

A tiger started running when it saw a deer running at uniform velocity 2m/s at 15m far from it ..if the tiger ran at acceleratio

Masteriza [31]

You need to set their position functions equal to one another and so for the time t when that is true. That is when the tiger and the deer are in the same place meaning the tiger catches the dear
Xdear= 2t+15 deer position function.
(I integrated the velocity function )
To get the Tigers position function you must integrate the acceleration twice. This becomes
Xtiger=t^2
Now t^2=2t+15
Time t is when the tiger catches the deer
t^2-2t-15=0
(t-5)(t+3)=0 factored
t=5s is the answer you use (t=-3 is a meaningless solution)

7 0

3 years ago

4. Two people each have a mass of 55 kg. They are both in an

mestny [16]

Answer:

350 N

Explanation:

F=ma

$f = force \\ m = mass \\ a = acceleration$

$m = 2(55kg) + 240kg \\ a = 1.0 \frac{m}{ {s}^{2} }$

Force = 350 Newtons

4 0

3 years ago

3. A model rocket is launched straight upward at 58.8 m/s.

kolezko [41]

Answer:

a). 6 seconds

b). 12 seconds

c). 176.4 meters

Explanation:

a). Equation to be applied to calculate the time taken by the rocket to reach at the peak height,

v = u - gt

where v = final velocity

u = initial velocity = 58.8 m per sec

g = gravitational pull = 9.8 m per sec²

t = duration of the flight

At the peak height,

v = 0

Therefore, 0 = 58.8 - (9.8)(t)

t = $\frac{58.8}{9.8}$

= 6 seconds

b). Total time of flight = 2(Time taken to go up)

= 2×6

= 12 sec

c). Formula to get the peak height is,

$h=ut-\frac{1}{2}gt^2$

h = (58.8)6 - $\frac{1}{2}(9.8)(6)^2$

= 352.8 - 176.4

= 176.4 meters

8 0

3 years ago