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2 years ago

Object A is 71 degrees and object B is 75 degrees how will thermal energy flow

Physics

1 answer:

Tasya [4]2 years ago

3 0

Given :

Object A is 71 degrees and object B is 75 degrees .

To Find :

How will thermal energy flow.

Solution :

We know, by law of thermodynamics thermal energy will flow from higher temperature to lower temperature.

So, in the given question energy will flow from object B from object A.

Hence, this is the required solution.

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If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no

Yakvenalex [24]

<u>Answer</u>

3.7 Km south

<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

(+1.7) + (-5.4) = 1.7 - 5.4

= - 3.7 Km But negative was towards south.

∴ Answer = 3.7 Km south.

6 0

3 years ago

How does 3rd class lever make our work easier

sineoko [7]

In a third class lever, the effort is located between the load and the fulcrum. If the fulcrum is closer to the load, then less effort is needed to move the load. If the fulcrum is closer to the effort, then the load will move a greater distance. ... These levers are useful for making precise movements.

7 0

3 years ago

Read 2 more answers

A child uses a rubber band to launch a bottle cap at an angle of 37.0° above the horizontal. The cap travels a horizontal distan

zavuch27 [327]

Answer:

Initial velocity will be 1.356 m/sec

Explanation:

Let the initial speed = u

Angle at which rubber band is launched = 37°

Horizontal component of initial velocity $u_x=ucos\Theta =ucos37^{\circ}=0.7986u$

Time is given as t = 1.20 sec

Distance in horizontal direction = 1.30 m

We know that distance = speed × time

So time $t=\frac{distance}{speed}$

$1.20=\frac{1.3}{0.7986u}$

$u=1.356m/sec$

So initial velocity will be 1.356 m/sec

3 0

3 years ago

Calculate the moment of inertia for each scenario: (a) An 80.0 kg skater is approximated as a cylinder with a 0.140 m radius. (b

Zina [86]

Answer:

a) the moment of inertia is 0.784 Kg*m²

b) the moment of inertia is with arms extended is 1.187 Kg*m²

c) the angular velocity in scenario (b) is 4.45 rad/s

Explanation:

The moment of inertia is calculated as

I = ∫ r² dm

since

I = Ix + Iy

and since the cylinder rotates around the y-axis then Iy=0 and

I = Ix = ∫ x² dm

if we assume the cylinder has constant density then

m = ρ * V = ρ * π R²*L = ρ * π x²*L

therefore

dm = 2ρπL* x dx

and

I = ∫ x² dm = ∫ x² 2ρπL* x dx = 2ρπL∫ x³ dx = 2ρπL (R⁴/4 - 0⁴/4) = ρπL R⁴ /2 = mR² /2

therefore

I skater = mR² /2 = 80 Kg * (0.140m)²/2 = 0.784 Kg*m²

b) since the arms can be seen as a thin rod

m = ρ * V = ρ * π R²*L = ρ * π R²*x

dm =ρ * π R² dx

I1 = ∫ x² dm = ∫ x² * ρ * π R² dx = ρ * π R²*∫ x² dx = ρ * π R²* ((L/2)³/3 - (-L/2)³/3)

= ρ * π R²*2*L³/24 = mL²/12

therefore

I skater 2 = I1 + I skater = mL²/12 + mR² /2= 8 Kg* (0.85m)²/12 +(80-8) Kg * (0.140m)²/2 = 1.187 Kg*m²

c) from angular momentum conservation

I s2 * ω s2 = I s1 * ω s1

thus

ω s2 = (I s1 / I s2 )* ω s1 /= (0.784 Kg*m²/1.187 Kg*m²) * 6.75 rad/s = 4.45 rad/s

4 0

3 years ago

In an internal combustion engine, air at atmostpheric

notsponge [240]

Answer:

1302 K or 1029 C

Explanation:

Air at atmospheric pressure has pressure of 1 atm

20 C = 20 + 273 = 293 K

Assume ideal gas, according to the ideal gas law:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Where P1, V1 and T1 are the pressure, volume and temperature of the gas before the compression and P2, V2 and T2 are the pressure, volume and temperature of the gas after the compression

$T_2 = T_1\frac{P_2V_2}{P_1V_1}$

Since the gas is compressed to 1/9 of its original volume, V2/V1 = 1/9:

$T_2 = 293\frac{40}{9} = 1302 K$ or 1029 C

8 0

3 years ago