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tekilochka [14]

3 years ago

6

For a given amount of gas at a constant temperature, the volume of a gas varies inversely with its pressure is a statement of __

______ law.
a. Charles's
b. Avogadro's
c. Curie's
d. Boyle's

1 answer:

Ksivusya [100]3 years ago

8 0

Answer:

d. Boyle's

Explanation:

Boyle's Law: States that the volume of a fixed mass of gas is inversely proportional proportional to its pressure, provided temperature remains constant.

Stating this mathematically. this implies that:

V∝1/P

V = k/P, Where k is the constant of proportionality

PV = k

P₁V₁ = P₂V₂

Where P₁ and P₂ are the initial and final pressure respectively, V₁ and V₂ are the the initial and final volume respectively.

Hence the right option is d. Boyle's

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The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s = 6/t2, where t is mea

ycow [4]

Answer:

Velocity of the particle at time t = a

$v(a)=-\frac{12}{a^3}$

Velocity of the particle at time t = 1

$v(1)=-12m/s$

Velocity of the particle at time t = 2

$v(2)=-1.5m/s$

Velocity of the particle at time t = 3

$v(3)=-0.44m/s$

Explanation:

Displacement,

$s(t)=\frac{6}{t^2}$

Velocity is given by

$v(t)=\frac{ds}{dt}=\frac{d}{dt}\left ( \frac{6}{t^2}\right )=-\frac{12}{t^3}$

Velocity of the particle at time t = a

$v(a)=-\frac{12}{a^3}$

Velocity of the particle at time t = 1

$v(1)=-\frac{12}{1^3}=-12m/s$

Velocity of the particle at time t = 2

$v(2)=-\frac{12}{2^3}=-1.5m/s$

Velocity of the particle at time t = 3

$v(3)=-\frac{12}{3^3}=-0.44m/s$

8 0

3 years ago

Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively

Ronch [10]

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

For car 1, we have

$d_1 = u_1 t + \frac{1}{2}a_1 t^2$

where

$u_1 = 7 m/s$ is the initial velocity of car 1

$a_1 = 0.4 m/s^2$ is the acceleration of car 1

So the equation can be rewritten as

$d_1 = 7t + 0.2t^2$

For car 2, we have

$d_2 = u_2 t + \frac{1}{2}a_2 t^2$

where

$u_2 = 4 m/s$ is the initial velocity of car 2

$a_2 = 0.5 m/s^2$ is the acceleration of car 2

So the equation can be rewritten as

$d_2= 5t + 0.25t^2$

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

$d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2$

And solving for t, we find

$2t - 0.05t^2= 0\\t(2-0.05t)=0$

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

$d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m$

Learn more about accelerated motion:

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7 0

4 years ago

A car of mass 960 kg is free-wheeling down an incline at a constant speed of 9.0 m s–1.

Orlov [11]

Need to draw it first.....
but there is no option here to draw .. why?
anyway , here the car is going with a uniform velocity..
so the acceleration of the car=0, a=0
so resistive force on the car=component along the inclined plane of the weight of the car=mg*cos(90-15)=960*9.8*cos 75=?

4 0

3 years ago

Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.

kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, $4.67\times 10^7$

(b) The number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

Explanation :

<u>For part (a) :</u>

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{3.82\text{ days}}$

$k=1.81\times 10^{-1}\text{ days}^{-1}$

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.81\times 10^{-1}\text{ days}^{-1}$

t = time passed by the sample = 12 days

a = initially number of radon atoms = $4.1\times 10^8$

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

$12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}$

$a-x=4.67\times 10^7$

Thus, the number of radon atoms will remain after 12 days is, $4.67\times 10^7$

<u>For part (b) :</u>

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = $(4.1\times 10^8)-(4.67\times 10^7)$

The number of radon nuclei have decayed = $3.6\times 10^8$

Thus, the number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

5 0

3 years ago

Is aluminum foil reflecting onto something conduction, convection, or radiation?

mojhsa [17]

I had the SAME problem, put down Radiation and it’s thermal/light.

4 0

3 years ago