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Lilit [14]
3 years ago
7

Suppose the energy required to freeze 0.250 kg of water were added to the same mass of water at an initial temperature of 1.0 °C

. What would be the final temperature of the water?
Physics
1 answer:
Eduardwww [97]3 years ago
6 0
The answer is a because it is
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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

8 0
3 years ago
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When you're out of breath, you feel dizzy, lung pain and maybe even nausea or side pain.
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It's time for Santa to deliver his presents. Assuming he delivers presents all over the world, he has to travel about 75,000,000
algol13
2,419,354.8387960000
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3 years ago
The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur
Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

           P₂ is the pressure in the hose

           ρ is the density of the fluid

           h₁ is the initial height

           h₂ is the final height

           v₁ is the initial velocity of the fluid

           v₂ is the final velocity of the fluid  and

           g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

    = (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

    = 5.92 × 10⁵ Pa

    = 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

#SPJ4

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What is the minimum diameter necessary for a radio telescope working at f=1×10^10 Hz to be able to separate two objects 1 deg ap
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