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3 years ago

3)

1 answer:

6 0

As a liquid is heated, its vapor pressure increases until the vapor pressure equals the pressure of the gas above it. ... In order to form vapor, the molecules of the liquid must overcome the forces of attraction between them. The temperature of a boiling liquid remains constant, even when more heat is added.

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In a lecture demonstration, a 3.0-m-long vertical string with ten bolts tied to it at equal intervals is dropped from the ceilin

sleet_krkn [62]

<span>First of all, the bolts wont hit the ground at equal time intervals because each bolt is constantly increasing. Therefore, the top will have different velocity with the bottom bolt at all times. this means that the bolts will speed up more as they are higher which will make the interval time shorter. therefore, the time will decrease</span>

8 0

4 years ago

(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and dia

stealth61 [152]

Answer:

7160.2812 s or 1.988 hours

Explanation:

m = Mass of person

R = Radius of Earth = $6.37\times 10^{6}\ m$

g = Acceleration due to gravity = 9.81 m/s²

$\omega$ = Angular speed

Force at equator would be

$F_e=m(g-\omega^2R)$

Force at pole

$F_p=mg$

From the question

$F_e=\dfrac{1}{2}F_p\\\Rightarrow m(g-\omega^2R)=\dfrac{1}{2}F_p\\\Rightarrow \omega=\sqrt{\dfrac{g}{2R}}$

Time period is given by

$T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours$

The rotational period of the planet is 7160.2812 s or 1.988 hours

5 0

3 years ago

Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori

LenaWriter [7]

Answer:

Explanation:

Sam mass=75kg

Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

Vf=√247.36

Vf=15.73m/s

3 0

4 years ago

I will give 14 points & make you the brainiest

Virty [35]

Answer:

B. silicate rocks and metals

8 0

3 years ago

A 60-kg skier starts from rest at the top of a 80-meter high practice slope (A). He uses his poles to propel himself forward, do

Romashka-Z-Leto [24]

Explanation:

Given data

Mass of skier m=65kg

Height of slope =80m

Potential energy possessed by the skier (positive work done) =12000J

Analysing his energy at half way point on the hill

Potential energy will be equal to kinetic energy

Applying the conservative principles of energy when the skier is at half way point in the hill

I.e

mgh= 1/2mv²

The masses will cancel out

gh=v²

Given g= 9.81m/s²

Let's solve for the velocity

v²=9.81*80

v=√784.8

v=28.0m/s

Now we can solve for the kinetic energy

KE= 1/2mv²

KE =1/2*60*28²

KE =47063.9/2

KE=23531.9J

Hence we can calculate the energy gained by the skier during his movement

Energy gained = KE-PE

= 23531.9-12000

= 11531.9J

3 0

4 years ago