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Ber [7]
3 years ago
15

A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

rcury in this process?
Physics
1 answer:
loris [4]3 years ago
5 0

To solve this exercise we will use the concept related to heat loss which is mathematically given as

Q = mC_p \Delta T

Where,

m = mass

C_p= Specific Heat

\Delta T = Change in temperature

Replacing with our values we have that

m = 0.1g

C_p = 139J/Kg\cdot K \rightarrow Specific heat of mercury

\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K

Replacing

Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J

Therefore the heat lost by mercury is 0.09J

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What do you measure when you measure an object’s mass?
Effectus [21]

Answer:

how much space it takes up in the world

Explanation:

1) Mass is a measurement of the amount of matter something contains, while Weight is the measurement of the pull of gravity on an object. 2) Mass is measured by using a balance comparing a known amount of matter to an unknown amount of matter. Weight is measured on a scale.

3 0
3 years ago
Each value in nature has a number part, called its____<br> and a dimension, or unit
saul85 [17]

Answer:

<u>Magnitude</u>

Explanation:

Each value in nature has a number part, called its magnitude and a dimension called its unit.

For example,

The length of an object is 10 cm. It means that 10 shows the magnitude of length and cm shows its unit.

7 0
3 years ago
If the mass of an object is 8 kg and its momentum is -80 kgm/s, what is its velocity?
Dimas [21]

An object's momentum is the product of its mass and its velocity:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

p = -80kg×m/s

m = 8kg

Plug in these values and solve for v:

-80 = 8v

v = -10m/s

Choice D

4 0
3 years ago
An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

F = (3v)^{2}

F = 9v^{2}

We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

F = (2v)^{2}

F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0
3 years ago
Consider an ideal monatomic gas of N particles with mass m in thermal equilibrium at a temperature T. The gas is contained in a
harina [27]

Answer:

K.E.=\dfrac{3}{2}KT

Explanation:

Given that

Number of particle =N

Equilibrium temperature= T

Side of cube = L

Gravitational acceleration =g

The kinetic energy of an atom  given as

K.E.=\dfrac{3}{2}KT

Where

Equilibrium temperature= T

Boltzmann constant =K

        K =1.380649×10−23 J/K

3 0
3 years ago
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