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2 years ago

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As a sample of water is heated, the kinetic energy of its molecules increases. Then, the energy is used to overcome the attracti

ons between the molecules so that they are no longer held close together. Which sections of the heating curve illustrate this process?

1 answer:

maks197457 [2]2 years ago

7 0

Answer: SECTION B (BOILING POINT)

Explanation: Boiling point (100°c)

This is the point in which the water molecules acquires sufficient kinetic energy to overcome the attraction binding them.(a point at which a liquid will boil).

When a liquid example water is heated to it's boiling point,there is a rapid vaporization,this is known as boiling.At this boiling point,the temperature of the vapour pressure of the liquid will be directly proportional to the pressure exerted on the liquid by the sorounding atmosphere.

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A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t

sertanlavr [38]

Explanation:

The given data is as follows.

C = $20 \times 10^{-6} F$

R = $100 \times 10^{3}$ ohm

$Q_{o} = 100 \times 10^{-6}$ C

Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

$Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

$13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

0.135 = $e^{\frac{-t}{2}}$

$e^{\frac{t}{2}} = \frac{1}{0.135}$

= 7.407

$\frac{t}{2} = ln (7.407)$

t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

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2 years ago

A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration

wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

$V=X/T$

First stage:

T1=5s

$v_{f} =v_{o} - at$

But, $v_{f} =0$ (decelerates to rest)

then: $a =v_{o} /t=0.3/5=0.06m/s^{2}$

on the other hand:

$x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m$

X1=75cm

Second stage:

T2=5s

$x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m$

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

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2 years ago

A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.

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guapka [62]

Answer:

Explanation:

Given two vectors as follows

E₁ = 13.5 i -12 j

E₂ = -7.4 i - 4.7 j

Resultant E = E₁ + E₂

= 13.5 i -12 j -7.4 i - 4.7 j

E = 6.1 i - 16.7 j

a ) X component of resultant = 6.1 N

b ) y component of resultant = -16.7 N

Magnitude of resultant = √ ( 6.1² + 16.7² )

= 17.75 N

d ) If θ be the required angle

tanθ = 16.7 / 6.1 = 2.73

θ = 70° .

counterclockwise = 360 - 70 = 290°

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zubka84 [21]

Answer:

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Explanation:

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