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Why is boiling water considered a physical change?

Liono4ka [1.6K]

Answer:

Boiling water is an example of a physical change and not a chemical change because the water vapor still has the same molecular structure as liquid water.

(credits to lumenlearning)

Explanation:

7 0

3 years ago

Read 2 more answers

A pitcher throws a curveball that reaches the catcher in 0.64 s. The ball curves because it is spinning at an average angular ve

myrzilka [38]

Answer:

$\Delta \theta=19.44\ rad$

Explanation:

We just have to calculate what angular displacement a ball with an average angular velocity of 290 rev/min experiments in 0.64s. By definition, angular velocity $\omega$ is the angular displacement $\Delta \theta$ divided by the time elapsed:

$\omega=\frac{\Delta \theta}{\Delta t}$

Since $1\ rev=2\pi \ rad$ and $1\ min=60\ s$ , we can covert:

$\omega=290\ rev/min=\frac{290\ rev}{min}(\frac{1\ min}{60s})(\frac{2\pi \ rad}{1\ rev})=30.37\ rad/s$

Where the terms between parenthesis are equal to 1, so they just change the units. Then for our values we have:

$\Delta \theta=\omega \Delta t=(30.37\ rad/s)(0.64s)=19.44\ rad$

6 0

4 years ago

How much heat is absorbed by a 75g iron skillet when it’s temperature rises from 5c to 20c?

alexira [117]

Answer: $Q=499.5 J$

Explanation:

The heat (thermal energy) absorbed by the iron skillet can be found using the following equation:

$Q=m.C.\Delta T$ (1)

Where:

$Q$ is the heat (absorbed)

$m=75 g$ is the mass of the element (iron in this case)

$C$ is the specific heat capacity of the material. In the case of iron is $C=0.444\frac{J}{g\°C}$

$\Delta T=T_{f}-T_{i}=20\°C - 5\°C= 15\°C$ is the variation in temperature

Knowing this, lets rewrite (1) with these values:

$Q=(75 g)(0.444\frac{J}{g\°C})(15\°C)$ (2)

Finally:

$Q=499.5 J$

6 0

3 years ago

The electric potential at a position located a distance of 20.7 mm from a positive point charge of 8.60×10-9C and 15.1 mm from a

max2010maxim [7]

Answer:

q2 = -4.35*10^-9C

Explanation:

In order to find the values of the second charge, you use the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

V: electric potential = 1.14 kV = 1.14*10^3 kV

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1: charge 1 = 8.60*10^-9 C

q2: charge 2 = ?

r1: distance to the first charge = 20.7mm = 20.7*10^-3 m

r2: distance to the second charge = 15.1mm

You solve the equation (1) for q2, and replace the values of the other parameters:

$q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C$

The values of the second charge is -4.35*10^-9C

8 0

3 years ago

A 1 kg ball and a 10 kg ball are dropped from a height of 10 m at the same time. In the absence of air resistance,

RideAnS [48]

Answer:

the two balls will hit the ground at the same time.

Explanation:

The time of dropping, in the following equation, is related to both the distance travel s and the gravitational acceleration g, which are the same for both ball (if we neglect air resistance), no matter what their mass are.

$s = gt^2/2$

$t^2 = 2s/g$

So the time it takes to drop 2 balls are the same. They will hit the ground at the same time.

3 0

4 years ago