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2 years ago

15

PLEASE HELP.You hit a hockey puck across an empty hockey rink. At first the puck moves quickly, but then it slows and comes to a

stop. What caused the hockey puck to stop?

2 answers:

saveliy_v [14]2 years ago

7 0

Hi there!

Two possible answers are air resistance and friction.

Friction is caused by the rubbing of the surface of the ground and the surface of the object. Although ice doesn't have much friction, it can still cause friction.

Air resistance is caused by friction between the air and the object. As the object moves along a surface, it collides into many air particles; thus, it slows down.

Hope this helps.
Have an awesome day! :)

Greeley [361]2 years ago

6 0

As far as it goes the slower it goes cause it is heavy and when its slowing down the ice makes it stop.

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Heat is added to a 2kg piece of ice at a rate of 793kW. How long will it take for ice to melt if it was initially 0?

Ede4ka [16]

Answer:

0.84 s

Explanation:

Step 1

Given information:

Mass of the ice (m) = 2.0 kg

Heat transfer rate (Q/T) = 793.0 kW

Latent heat of fusion of ice (Lf) = 334 kJ/kg

$\frac{Q}{T} = \frac{mLf}{T}$

Substituting the corresponding values we have:

$793.0 kW= \frac{2.0kg(334 kJ/kg)}{T} \\ T = \frac{2.0kg(334 kJ/kg)}{793.0kW} = \frac{668kJ}{793kW} \\ = 0.84s$

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2 years ago

An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 a

sveticcg [70]

Answer:

<h2>2.22 kPa</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222$

We have the final answer as

<h3>2.22 kPa</h3>

Hope this helps you

5 0

2 years ago

In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot

Fittoniya [83]

Answer:

$Q=7.9\times 10^{-10}\ C$

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

$A=5\times 10^{-3}\ m^2$

We Capacitance given as

For air

$C_1=\dfrac{\varepsilon _oA}{d}$

$C_1=\dfrac{\varepsilon _oA}{d}$

$C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}$

$C_1=2.2\times 10^{-11}\ F$

$C_2=\dfrac{K\varepsilon _oA}{d}$

$C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F$

$C_2=6.6\times 10^{-11}\ F$

Net capacitance

C=C₁+C₂

$C=8.8\times 10^{-11}\ F$

We know that charge Q given as

Q= C V

$Q=12\times 6.6\times 10^{-11}\ C$

$Q=7.9\times 10^{-10}\ C$

6 0

2 years ago

A 12.0 kg rock slides off the edge of a bridge and falls into the water 25.0 meters below. what is the kinetic energy of the roc

Pavel [41]

During the fall, all the initial potential energy of the rock
$U=mgh$
has converted into kinetic energy of motion
$K= \frac{1}{2}mv^2$
where h is the initial height of the rock, m its mass, and v its velocity just before hitting the water. So, for energy conservation, we have
$U=K$
and so we can find the value of K, the kinetic energy of the rock just before hitting the ground:
$K=U=mgh = (12.0 kg)(9.81 m/s^2)(25.0 m)=2943 J$

7 0

2 years ago

Se deja caer un objeto desde la cornisa de una casa la cual tarda en llegar al suelo 645/500 segundos. ¿Cúal es la altura de la

salantis [7]

8.15m. La altura de la casa de la cual se deja caer un objeto que tarda en caer $\frac{645}{500}s$ es de 8.15m.

La clave para resolver este problema es mediante caída libre, la velocidad inicial del objeto es 0 por lo que podemos calcular la altura h de la casa mediante la relación $h= \frac{gt^{2}}{2}$ .

Conocemos la gravedad $g=9.8\frac{m}{s^{2}}$ y el tiempo en que tarda en caer el objeto $t=\frac{645}{500}s$ , sustituyendo los valores tenemos que:

$h=\frac{(9.8\frac{m}{s^{2}})(\frac{645}{500}s)^{2}}{2}=8.15m$

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2 years ago