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azamat
3 years ago
7

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

drop?. . .
a.The molecules in the water spread apart. . .
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 increases
Physics
2 answers:
katovenus [111]3 years ago
4 0
The correct answer should be c.The kinetic energy of the water molecules decreases.

If the temperature drops that means that the molecules are coming together. If the temperature rises then it means that the molecules are spreading. If the kinetic energy falls down that means that they are slower which means that they are cooler.
Anna007 [38]3 years ago
4 0

Answer: The correct answer is (c).

Explanation:

Thermometer is a device which measures the temperature.

Temperature is a measures of the average kinetic energy of the particles in the substance. It measures how the hot or cold an object is.

In the given problem, a thermometer is placed in water in order to measure the water’s temperature. When the kinetic energy of the water molecules decreases then the thermometer shows the drop in the temperature.

When the kinetic energy of the water molecules increases then the thermometer shows the rise in the temperature.

Therefore, the correct option is (c).

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Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the
Nina [5.8K]

Answer:

a= 92. 13 m/s²

Explanation:

Given that

Amplitude ,A= 0.165 m

The maximum speed ,V(max) = 3.9 m/s

We know that maximum velocity in the SHM  given as

V(max)  = ω A

ω=Angular speed

A=Amplitude

\omega =\dfrac{3.9}{0.165}\ rad/s

ω=23.63 rad/s

The maximum acceleration given as

a = ω² A

a= (23.63)² x 0.165 m/s²

a= 92. 13 m/s²

Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².

5 0
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Pushing a heavy sofa across a carpeted room is an example of
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Force is the answer so
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alexandr402 [8]

Answer:

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2 years ago
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For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec
zavuch27 [327]

Answer: b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

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3 years ago
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