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2 years ago

15

In what way are liquids different from solids?.

1 answer:

puteri [66]2 years ago

7 0

Answer:

solid molecules are packed closely and liquid molecules are losely packed

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A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as

Taya2010 [7]

Answer:

$w = - 508.53$ joules

$q = - 3091.47$ joules

Explanation:

Let us convert the time in hours into seconds

$0.010* 3600\\= 36$

Change in internal energy

$\delta E = p * \delta t$

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

$\delta E = - 100 * 36\\$

$\delta E = - 3600$ Joules

Amount of work done by the system

$w = - P * \delta V$

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

$w = - 1 * ( 5.92 -0.90)\\$

$w = -5.02$ liter-atmospheres

Work done in Joules

$- 5.02 * 101.3\\$ $= 508.53$ Joules

$q = \delta E - w\\$

Substituting the given values we get -

$q = - 3600 - (-508.53)\\q = - 3091.47$

Thus

$w = - 508.53$ joules

$q = - 3091.47$ joules

7 0

3 years ago

Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

nalin [4]

Answer:

$13 W/m^2$

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

$I \propto \frac{1}{r^2}$

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

$\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2}$ (1)

where in this problem, we have:

$I_1 = 1300 W/m^2$ apparent brightness at a distance $r_1$ , where

$r_1 = 150$ million km

We want to estimate the apparent brightness at $r_2$ , where $r_2$ is ten times $r_1$ , so

$r_2 = 10 r_1$

Re-arranging eq.(1), we find $I_2$ :

$I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2$

5 0

3 years ago

How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead

lara [203]

Answer:

Heat required = mass× latent heat Q = 0.15 × 871 ×

3 0

3 years ago

As a result of fringing effect :

PSYCHO15rus [73]

i thinks answers is gap relutance increases linearly with magnetic density

5 0

3 years ago

Solenoid 2 has twice the radius and six times the number of turns per unit length as solenoid 1. The ratio of the magnetic field

Xelga [282]

Answer:

6

Explanation:

The magnetic field inside a solenoid is given by the following formula:

$B = \mu_{0}nI$

where,

B = Magnetic Field Inside Solenoid

μ₀ = permittivity of free space

n = No. of turns per unit length

I = Current Passing through Solenoid

For Solenoid 1:

$B_{1} = \mu_{0}n_{1}I ------------------- equation 1$

For Solenoid 2:

n₂ = 6n₁

Therefore,

$B_{1} = \mu_{0}n_{2}I\\B_{1} = 6\mu_{0}n_{1}I ----------------- equation 2$

Diving equation 1 and equation 2:

$\frac{B_{2}}{B_{1}} = \frac{6\mu_{0}nI}{\mu_{0}nI}\\\\\frac{B_{2}}{B_{1}} = 6$

Hence, the correct option is:

<u>6</u>

8 0

2 years ago