All categories

3 years ago

Why is it important to use constants in an experiment?

1 answer:

5 0

<span>In order for the results to be valid, the dependent variable can only be affected by the independent variable, so somethings need to be kept constant. The things that need to be kept constant are called controlled variables.</span>

You might be interested in

A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th

kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

$P_1, V_1, T_1 & P_2, V_2, T_2$

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

$-Q=W+U_1----1$

$Q=-W+U_2-----2$

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

$U_1 & U_2$ change in internal Energy of gas

Thus from 1 & 2 we can say that

$U_1+U_2=0$

$n_1c_v(T-T_1)+n_2c_v(T-T_2)=0$

$T(n_1+n_2)=n_1T_1+n_2T_2$

$T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}$

where $n_1=\frac{P_1V_1}{RT_1}$

$n_2=\frac{P_2V_2}{RT_2}$

$T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}$

$T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}$

and $P=\frac{P_1V_1+P_2V_2}{V_1+V_2}$

6 0

3 years ago

Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

3 years ago

On a trolley ride around an amusement park, a child travelled from one signpost to a second signpost at a constant speed of 125

Tomtit [17]

Answer:

Explanation:

Speed given = 125 m /min

125 /60 m /s

In 450 second it will travel

= 450 x 125 / 60

=937.5 m.

As the distance is covered in less than 450 seconds , The distance must be less than 937.5 m

In 400 seconds , it will travel

= 400 x 125 / 60

833.33 m

Since the distance is covered in more than 400 seconds , the distance must be more than ie 833.33 .

Hence the distance covered is more than .833 m but less than 937.5

In either case these distance are more than .8 km .

5 0

2 years ago

Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the i

ELEN [110]

Answer:

Pls see attached file

Explanation:

5 0

3 years ago

When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache

Masteriza [31]

Answer:

b) True. the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

fr - w = ma

Where the friction force has some form of type.

fr = G v + H v²

Let's replace

(G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

fr - w = 0

fr = mg

(G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

G v = mg

v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0

3 years ago