All categories

3 years ago

According to gay-lussac’s law: select one:

Physics

2 answers:

DaniilM [7]3 years ago

8 0

Gay-Lussac's Law states
P1 / T1 = P2 / T2
So the answer is b

svlad2 [7]3 years ago

4 0

Answer

Explanation:

Just did the assignment.

You might be interested in

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh

Svetradugi [14.3K]

Answer:

The time needed is $T = 16.8 s$

Explanation:

From the question we are told that

The magnitude of the stimulated acceleration due gravity is $a = 0.5 g$

The diameter of the spaceship is $d = 35m$

Generally the force acting on the spaceship is

$F = ma$

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

$F = \frac{mv^2}{r}$

Thus

$ma = \frac{mv^2}{r}$

=> $\frac{v^2}{r} = a$

Generally the speed of this spaceship is mathematically represented as

$v = \frac{2 \pi}{T}$

=> $v^2 = [\frac{2\pi}{T}] ^2$

=> $\frac{\frac{4\pi^2 r^2}{T^2} }{r} = 0.5g$

=> $\frac{4 \pi^2 r }{T^2} = 0.5 g$

=> $T = \sqrt{ \frac{4\pi^2 r}{0.5g}}$

substituting values

$T = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}$

$T = 16.8 s$

4 0

3 years ago

Read 2 more answers

Where is the energy in a glucose molecule stored?

Semmy [17]

Answer:

Energy is stored in the bonds between atoms

4 0

3 years ago

When you hit a .27 kg volleyball the contact time is 50 ms and the average force is 125 N. If you serve the volleyball (from res

yulyashka [42]

(A) P(v) = 0.135v

(B) P(h) = 0.234v

<u>Explanation:</u>

Given-

Mass of the ball, m = 0.27kg

Force, F = 125N

angle of projection, θ = 30°

Let v be the velocity of the ball.

A) vertical component of the momentum of the volleyball

We know,

P(vertical) = mvsinθ

P(V) = 0.27 X v X sin 30°

P(V) = 0.27 X v X 0.5

P(V) = 0.135v

B) horizontal component of the momentum of the volleyball

We know,

P(Horizontal) = mvcosθ

P(h) = 0.27 X v X cos 30°

P(h) = 0.27 X v X 0.866

P(h) = 0.234v

8 0

3 years ago

A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o

Marrrta [24]

Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

Initial energy

$KE_i=\dfrac{1}{2}mu^2$

$KE_i=\dfrac{1}{2}0.3\times 10^2$

KEi= 15 J

Final kinetic energy

$KE_f=\dfrac{1}{2}mv^2$

$KE_f=\dfrac{1}{2}0.3\times 5^2$

KEf=3.75 J

The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

The minimum value to complete the circular arc

$V=\sqrt{r.g}$

Now by putting the values

$V=\sqrt{0.8\times 10}$

V= 2.82 m/s

So kinetic energy KE

$KE=\dfrac{1}{2}mV^2$

$KE=\dfrac{1}{2}0.3\times 2.82^2$

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

= 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

4 0

3 years ago

Take schlatts love uwu (i cant spell)

7nadin3 [17]

thank you so much for the schlatt

8 0

3 years ago