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Elan Coil [88]
3 years ago
7

Carmen is heating some water and trying to measure the temperature of water using a Celsius thermometer. Which measurement can s

he expect once the water begins to boil?
32°C
100° C
212°C
373°C

Physics
2 answers:
Aleksandr-060686 [28]3 years ago
6 0

Answer:

option b) 100°

Explanation:

andrew-mc [135]3 years ago
4 0

<u>100° C</u> she can expect once the water begins to boil.

<u>Option: B</u>

<u>Explanation:</u>

The boiling point for water at 1 pressure atmosphere of sea level is 212 ° F or 100 ° C. That value isn't a fixed. Water's boiling point is dependent on the ambient pressure, which varies based on elevation. At a lower temperature, water boils as one gains altitude like getting higher on a hill, and boils at a higher temperature if one increases the atmospheric pressure of returning to or below sea level.

It also relies upon the water's purity. Water containing contaminants like salted water boils at a level higher than pure water. This effect is called acceleration of the boiling point and is one of the material's colligative features.

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3 years ago
A block sliding across a level surface has a mass of 2.5 kg and a mechanical energy of 20 joules. What is its velocity?
tia_tia [17]
Hi!

The energy of the block is 4 m/s

To calculate this, you need to use the equation for kinetic energy. The block is sliding (i.e. it's moving). If the object is sliding across a level surface, the only energy it has is kinetic energy, because there is no change in potential energy (which changes with height). So, the mechanical energy will be pure kinetic energy. The equation is the following, derived from the expression for kinetic energy:

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

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v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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Answer:

c

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