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3 years ago

7. 10 Kg of water at 90°C is cooled by mixing 20kg of water at 20°C. What

Physics

1 answer:

BlackZzzverrR [31]3 years ago

3 0

Answer:

Please mark me as Brilliant

Explanation:

I shall solve both one by one

Case 1: 1 kg of water at 10° C

Let the final temperature be T° C , where T°C < 30°C.

Heat give out by 4.4 kg of water at 30° C to come down to T°C = 4400 g × 1 calorie/g/°C × (30- T)° C = (4400× 30 - 4400 T) calorie

Heat gained to raise temperature of 1kg of water at 10° to T° = 1000 g × 1 calorie/g/°C × (T -10)° C = 1000 T - 10,000) calorie

Heat gained = Heat lost

1000 T - 10,000 = 1,32,000 - 4400 T; ==> 54 00T = 1.42,000; T = 1,42,000/5400 = 26.3° C.

In case it is 1 kg of water at 10° C mixed with 4.4 kg of water at 30° C, the final temperature of the mixture would be T = 26.3° C.

Case 2: 1 kg of ice.

Ice is essentially at its melting point/freezing point ie at 0° C.

Let the temperature of mixture = T° C

Heat required to melt 1 kg (=1000 g of ice) at 0° C to water at 0° C = 1000 g × 80 calorie/g = 80,000 calorie

Heat require to raise temperature of 1000 g of water at 0° C to water at T° C = 1000g × 1 calorie/g/°C× (T -0)° C = 1000 T calorie

Heat gained = 80,000 + 1000 T

Heat lost by 4.4 kg of water at 30° C to cool to T° C = 4400 g × 1 calorie/g /° C × ( 30 - T)° C = 1,32,000 - 4400 T

Heat gained = Heat lost

80,000 + 1,000 T = 1,32,000 -4400 T

5400 T = 52,000/5400; ==> T = 9.63°C.

In case it is ice, the temperature of the mixture is T = 9.63° C.

Added: around 2 pm

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PLEASE PROVIDE EXPLANATION.<br><br> THANK YOU!!

Ksju [112]

Answer:

11,000 kg

(a) 11.2 m/s

(b) 1.6 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(2200 kg) (60.0 km/h) + m (0 km/h) = (2200 kg) (10 km/h) + m (10 km/h)

132,000 kg km/h = 22,000 kg km/h + m (10 km/h)

110,000 kg km/h = m (10 km/h)

m = 11,000 kg

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(m) (-v) + (2m) (5v) = (m) (v₁) + (2m) (v₂)

-mv + 10mv = m v₁ + 2m v₂

9mv = m (v₁ + 2 v₂)

9v = v₁ + 2 v₂

Since the collision is elastic, kinetic energy is also conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(m) (-v)² + (2m) (5v)² = m v₁² + (2m) v₂²

mv² + 50mv² = m v₁² + 2m v₂²

51mv² = m (v₁² + 2 v₂²)

51v² = v₁² + 2 v₂²

We know v = 1.60 m/s. So the two equations are:

14.4 = v₁ + 2 v₂

130.56 = v₁² + 2 v₂²

Solve the system of equations using substitution.

130.56 = (14.4 − 2 v₂)² + 2 v₂²

130.56 = 207.36 − 57.6 v₂ + 4 v₂² + 2 v₂²

0 = 6 v₂² − 57.6 v₂ + 76.8

0 = v₂² − 9.6 v₂ + 12.8

v₂ = [ 9.6 ± √(9.6² − 4(1)(12.8)) ] / 2(1)

v₂ = 1.6 or 8

If v₂ = 1.6 m/s, then v₁ = 14.4 − 2 v₂ = 11.2 m/s.

If v₂ = 8 m/s, then v₁ = 14.4 − 2 v₂ = -1.6 m/s.

We know v₁ can't be -1.6 m/s, since that would mean puck A didn't change speeds after the collision. Therefore, v₁ = 11.2 m/s and v₂ = 1.6 m/s.

8 0

3 years ago

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 85.0 m/s2

sergey [27]

Answer:

Maximum height attained by the model rocket is 2172.87 m

Explanation:

Given,

Initial speed of the model rocket = u = 0
acceleration of the model rocket = $a\ =\ 85.0 m/s^2$
time during the acceleration = t = 2.30 s

We have to consider the whole motion into two parts

In first part the rocket is moving with an acceleration of a = 85.0 $m/s^2$ for the time t = 2.30 s before the fuel abruptly runs out.

Let $s_1$ be the height attained by the rocket during this time intervel,

$s_1\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s_1\ =\ 0\ +\ 0.5\times 85\times 2.30^2\\\Rightarrow s_1\ =\ 224.825\ m$

And Final velocity at that point be v

$\therefore v\ =\ u\ +\ at\\\Rightarrow v\ =\ 0\ +\ 85.0\times 2.3\\\Rightarrow v\ =\ 195.5\ m/s.$

Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,

Let ' $s_2$ ' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,

At that insitant,

initial velocity of the rocket = v = 195.5 m/s.

a = $-g\ =\ -9.81\ m/s^2$
Final velocity of the rocket at the maximum altitude = $v_f\ =\ 0$

From the kinematics,

$v^2\ =\ u^2\ +\ 2as\\\Rightarrow 0\ =\ u^2\ -\ 2gs_2\\\Rightarrow s_2\ =\ \dfrac{u^2}{2g}\\\Rightarrow s_2\ =\ \dfrac{195.5^2}{2\times 9.81}\\\Rightarrow s_2\ =\ 1948.02\ m$

Hence the maximum altitude attained by the rocket from the ground is

$s\ =\ s_1\ +\ s_2\ =\ 224.85\ +\ 1948.02\ =\ 2172.87\ m$

6 0

3 years ago

a 5-kg fish swimming at 1 m/s swallows an absent minded 1-kg fish at rest. What is the speed of the large fish immediately afyer

storchak [24]

Answer:

In first case speed of fish will be $\frac{5}{6}$ m/s.

In the second case the speed of fish $\frac{1}{6}$ m/s

Explanation:

Given data :-

Mass of bigger fish ( m₁ ) = 5 kg.

Mass of small fish ( m₂ ) = 1 kg.

Speed of large fish ( v₁ ) = 1 m/s

Mass of bigger fish after eating smaller one = 5 + 1 = 6 kg.

Case - 1

Momentum of bigger fish before eating the smaller fish = m₁* v₁ = 5 * 1 = 5 kg.m/s

Momentum of bigger fish after eating the larger fish = ( m₁ + m₂)*v

v = speed of bigger fish immediately after lunch.

Using the conservation of momentum.

m₁* v₁ = ( m₁ + m₂)*v

5 = 6 * v

v = $\frac{5}{6}$ m/s.

Case -2

Speed of small fish = 4 m/s

Momentum of bigger fish before lunch = 5 kg.m/s

Momentum of smaller fish before lunch = 4*1 = 4 kg.m/s

Net momentum before lunch = 5 - 4 = 1 kg.m/s

Momentum of bigger fish after eating the larger fish = 6 * V

Using the conservation of momentum.

1 = 6 * V

V = $\frac{1}{6}$ m/s.

3 0

3 years ago

Which state of matter has the greatest amount of kinetic energy?

zvonat [6]

Answer:

A pure substance in the gaseous state contains more energy than in the liquid state, which in turn contains more energy than in the solid state. Particles has the highest kinetic energy when they are in the gaseous state.

Explanation:

Sana makatulong.

3 0

3 years ago

An 80-g particle moving with an initial speed of 50 m/s in the positive x direction strikes and sticks to a 60-g particle moving

liubo4ka [24]

The collision is a form of inelastic collision because the it forms a single mass after is collides. So it can be solve by momentum balance

( 0.08 kg * 50 m/s ) + ( 0.06 kg * 50 m/s) = ( 0.08 + 0.06 kg ) v

V = 50 m/s

So the kinetic energy lost is

KE = 0.5 (50 m/s)^2) *( 0.14 – 0.08kg )

KE = 75 J

8 0

3 years ago