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GarryVolchara [31]
3 years ago
11

Explain how we were able to boil water at room temperature

Physics
1 answer:
alekssr [168]3 years ago
5 0

Answer:

put water at room temperature into a vacuum chamber and begin removing the air. Eventually, the boiling temperature will fall below the water temperature and boiling will begin without heating.

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Relationship between education and income
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To summarise the findings, in 2012, there is a significant and positive relationship between higher education level and income (i.e., higher education qualifications lead to higher income). ... Next, we examine the incomes of between bachelor's degree and graduate degree holders.
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3 years ago
You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l
Ugo [173]

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

E_p = E_{kv} + E_{k\omega}

mgh = mv^2/2 + I\omega^2/2

where m is the mass, I = mr^2/2 is the moments of inertia of the solid cylinder \omega = v / r is the angular speed of the cylinder

mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}

mgh = mv^2/2 + mv^2/4 = 3mv^2/4

h = 3v^2/(4g)

So if you plot a liner chart of h vs v^2 and get a slope of 6.42 then that means 3/(4g) = 6.42 so g = 6.42*4/3 = 8.56 m/s^2

The gravitational acceleration on this planet is 8.56 m/s2

3 0
4 years ago
An astronaut decides to perform an experiment to monitor how much weight he loses during his stay on the International Space Sta
kumpel [21]

Answer:

The first one is: His weight on the Earth before take-off and the weight after take-off back on Earth once he gets back should be recorded as his Independent variable and his dependent variable.

The second one is: If he gained the weight back that he had lost while on the trip then you should disregard them unless that was the weight he was when he weighed himself after he got back.

The Third one is: The mass of an object is the amount of matter it contains, regardless of its volume or any forces acting on it. … Gravity is a force that attracts objects toward the Earth. The weight of the object is defined as the force caused by gravity on a mass.

Explanation:

I took  the quiz earlier. Hope this Helps you.

4 0
3 years ago
Describing motion using words instead of numbers is called
GrogVix [38]
The answer is <span>Qualitative</span>
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3 years ago
Read 2 more answers
The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21
frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: <em>The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is </em>h(t) = 3+21.5\cdot t-4.9\cdot t^{2}<em>after t seconds. (Round your answers to two decimal places.) </em><em>(a)</em><em> Find the velocity after 2 seconds and after 4 seconds, </em><em>(b)</em><em> When does the projectile reach its maximum height? </em><em>(c)</em><em> What is the maximum height? </em><em>(d)</em><em> When does it hit the ground? </em><em>(e)</em><em> With what velocity does it hits the ground?</em>

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

v(t) = 21.5-9.8\cdot t (Eq. 1)

Where v(t) is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

v(2) = 21.5-9.8\cdot (2)

v(2) = 1.9\,\frac{m}{s}

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

v(4) = 21.5-9.8\cdot (4)

v(4) = -17.7\,\frac{m}{s}

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for t:

21.5-9.81\cdot t = 0

t = 2.192\,s

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}

h (2.192) = 26.584\,m

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for t. That is:

3+21.5\cdot t-4.9\cdot t^{2} = 0

Roots are found by means of Quadratic Formula:

t_{1}\approx 4.523\,s and t_{2}\approx -0.135\,s

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

v(4.523) = 21.5-9.81\cdot (4.523)

v(4.523) = -22.871\,\frac{m}{s}

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

5 0
4 years ago
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