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2 years ago

When cold milk

Physics

1 answer:

monitta2 years ago

7 0

Answer:

See explanation below.

Explanation:

The internal energy (kinetic energy) of the milk particles increases thus warming the milk up, while the internal energy of the coffee particles reduce their initial average speed as they are imparting energy to the slow milk particles.

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If you could travel at the speed of light, how long would it take to travel from our solar

Illusion [34]

Answer:

4.689 years

Explanation:

6 0

3 years ago

Which best describes most covalent compounds?<br> O soft<br> O brittle<br> 3<br> O cold<br> O warm

LUCKY_DIMON [66]

Answer: Brittle

Explanation:

took the test and I chose Soft, Soft is the wrong answer don't choose it. The CORRECT ANSWER IS BRITTLE

7 0

3 years ago

Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims

777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force $F = 12.3$ N

Displacement $d = 10.2$ m

Mass of rocket $m = 0.663$ Kg

From work energy theorem,

$W = \Delta K$

$W_{t} - Wd - W_{g} = KE$

Where $W_{t} =$ thrust work $W_{g} =$ gravitational work

$KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2$

$KE = 59.2 -Wd$

After cutoff kinetic energy is converted into potential energy,

$KE = Wd' + mg\Delta h$

Put value of KE

$59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5$

Work done by drag force is given by,

$Wd'+Wd = 59.2 -29.23$

$= 29.96$ J

Therefore, the work done by the drag force is given by 29.96 J

5 0

3 years ago

A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat

Dafna11 [192]

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is $\sqrt{3^2+(-5)^2}$ = $\sqrt{9+25}$ = $\sqrt{34}$

3 0

3 years ago

A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Natasha2012 [34]

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0

3 years ago