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const2013 [10]
3 years ago
13

DatePageWhat are the advantages of alcoholthermometric liquid?​

Physics
1 answer:
Leya [2.2K]3 years ago
3 0

Answer:

low freezing point. high vapour pressure.

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>WILL</em><em> </em><em>HELP</em><em> </em><em>U</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>

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a car with a mass of 100 kg is stopped on the side of the road after getting a flat tire. the two people that were riding in the
notsponge [240]

Answer:

Please mark me brainliest and thank me and rate me

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3 years ago
Why can people float effortlessly on the dead sea?
ahrayia [7]
There water has a lot of salt in it, so it would be denser. People float better when the water is denser.
5 0
3 years ago
A _____________ is a system that uses vertical posts which are separated to support a horizontal beam. A. Truss c. Post-and-lint
IgorC [24]

A Post-and-lintel is a system that uses vertical posts which are separated to

support a horizontal beam.

A Post-and-lintel system is commonly used in architecture and involves the

system where horizontal structures are held by vertical ones with the

presence of spaces between them.

In this scenario, we were told the system uses vertical posts which are

separated to support a horizontal beam which makes Post-and-lintel the

most appropriate choice.

Read more on brainly.com/question/8777125

3 0
2 years ago
a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(
horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.    

Explanation:

We can find the velocity of the skier by energy conservation:

E_{1} = E_{2}

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}    (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²  

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}  

v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s    

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!  

6 0
2 years ago
What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x
lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, U_{electric,b} - U_{electric,a} as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2×10^{-6} J

(b) KE = 2×10^{-6} J

Explanation: <u>Potential</u> <u>Energy</u> (U) is the amount of work done due to its position or condition and its unit is Joule (J). <u>Kinetic</u> <u>Energy</u> (KE) is the ability to do work by virtue of velocity and the unit is also (J). <u>Mechanical</u> <u>Energy</u> is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2×10^{3} N/C

q = 2×10^{-9} C

d = 50 - (-30) = 80×10^{-2} = 8×10^{-1}m

ΔU = U_{electric,b} - U_{electric,a} = Eqd

U_{electric,b} - U_{electric,a} = 2×10^{3} .  2×10^{-9} . 8×10^{-1}

ΔU = 3.2×10^{-6} J

(b) Mechanical Energy is constant, so:

KE_{i} + U_{i} = KE_{f} + U_{f}

Since the initial position is zero and there is no initial kinetic energy:

KE_{f} = - U{f}

KE_{f} = - (2×10^{3}. 2×10^{-9} . 5×10^{-1})

KE_{f} = - 2.10^{-6} J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

8 0
3 years ago
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