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Lostsunrise [7]
3 years ago
6

A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to

rise?
A. The molecules in the water move closer together.
B. The molecules in the thermometer’s liquid spread apart.
C. The kinetic energy of the water molecules decreases.
D. The kinetic energy of the thermometer’s liquid molecules decreases.
Physics
1 answer:
mart [117]3 years ago
7 0

Answer:

Option B

The molecules in thermometer spread apart

Explanation:

The mercury within the thermometer spread apart leading to an increase in temperature reading in the thermometer. The spreading apart of mercury molecules is caused by an increase in kinetic energy of molecules and with time the increase in kinetic energy decreases making the reading to stop increasing hence constant.

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Answer:

A radio telescope helped the astronomers discover the CMB.

Explanation:

  • Penzias and Wilson while experimenting with a radio telescope in 1964, accidentally discovered the radiation that exists universally also known as the CMB.
  • This was used to support the "Big Bang Theory" and not the "Steady State Theory"
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3 years ago
Which type of energy is commonly referred to as kinetic energy?
aksik [14]
Kinetic Energy is movement energy (most simplistic way I can put it) so its motion. 
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3 years ago
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Which quantity can be calculated using Newton’s second law of motion?
DiKsa [7]

Newton's 2nd law of motion:

          Net Force  =  (mass) x (acceleration) .

The law shows the relationship among an object's mass
and acceleration, and the net force acting on it.

If you know any two of the quantities in the formula,
the law can be used to calculate the third one.
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3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

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What you do is, multiply 16.0 and 12.4 together. then multiply that by 40a
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