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FinnZ [79.3K]
3 years ago
14

Plz do all of it i will give brainlest and thanks to best answer plz do it right

Physics
1 answer:
Sladkaya [172]3 years ago
8 0

Answer: The answer is B: Convection  

Explanation: Convection transfers heat through a movement of fluids but in this case its through air hope this helps :)

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a/ Compute the quantity of heat released by 25.0 g of steam initially at 100.0oC, when it is cooled to 34.0°C and by 25.0 g of w
Svetach [21]

Answer:

When put into steam

Explanation:

When a certain amount of steam at boiling temperature condenses (turning into water), the amount of heat released is

Q_1 = m\lambda_v

where in this case

m = 25.0 g = 0.025 kg is the mass of steam at 100.0°C

\lambda_v=2.256\cdot 10^6 J/kg is the latent heat of vaporization of water

So,

Q_1=(0.025)(2.256\cdot 10^6)=56400 J

Instead, the amount of heat released when the water at 100.0°C is cooled down to 34.0°C is given by

Q_2=mC\Delta T

where

m = 25.0 g = 0.025 kg is the mass of water

C=4.19\cdot 10^4 J/kg K is the specific heat of water

\Delta T=100-34=66^{\circ}C is the change in temperature

Therefore,

Q_2=(0.025)(4.19\cdot 10^3)(66)=6913 J

Since Q_1>Q_2, we can say that your hand will burn more in the first case.

5 0
3 years ago
Why cant body builders swim please give reasons.
Ronch [10]
The main reason is becouse these people are over 200 pounds and no mater how strong you are arms arnt ment to keep you up thanks.
8 0
3 years ago
Read 2 more answers
A 1400 kg car driving at 25 m/s slams on its brakes. The coefficient of kinetic friction between the tires and the road is 0.7.
tamaranim1 [39]

The acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

The given parameters;

  • mass of the car, m = 1400 kg
  • Initial velocity of the car, u = 25 m/s
  • coefficient of kinetic friction, μ = 0.7

The acceleration of the car is calculated as follows;

a = μg

a = 0.7 x 9.8

a = 6.86 m/s²

The time taken for the car to stop is calculated by using Newton's second law of motion;

F = ma

F = \frac{mv}{t} \\\\ma = \frac{mv}{t}\\\\a = \frac{v}{t} \\\\t = \frac{v}{a} \\\\t = \frac{25}{6.86} \\\\t = 3.64 \ s

Thus, the acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

Learn more here:brainly.com/question/19887955

5 0
3 years ago
A team of engineers is working together to design a new airplane. The team members each live in different cities. For which purp
Ierofanga [76]

Answer:

The internet is most useful to them because they use it to communicate.

Explanation:

If I were to send a message to my brother in Florida, through the internet, while I'm in Pennsylvania he would get it in minutes. On the other hand if I were going to meet him and then explain what I wanted to tell him in person it would take a much longer time.  

8 0
3 years ago
A 14.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initially at rest on a horizontal surface. The clay s
ahrayia [7]

Answer:

86.53 m/s

Explanation:

Given:

Mass of clay (m) = 14.0 g = 0.014 kg

Mass of block (M) = 110 g = 0.110 kg

Initial speed of block (U) = 0 m/s

Sliding distance (d) = 7.50 m

Coefficient of friction between block and surface (μ) = 0.650

Let the initial speed of clay be 'u' and speed of clay and block just after collision be 'v'.

Now, momentum is conserved just before and just after collision.

Momentum just before collision = mu + 0 = mu

Momentum just after collision = (m + M)v

Therefore, mu=(M+m)v --------- (1)

Now, using newton's second law and we find the acceleration of the system.

The frictional force is given as:

f=\mu mg=-ma\\\\a=-\mu g

Now, using equation of motion, we can find the velocity just after collision.

0^2=v^2+2ad\\\\v=\sqrt{-2ad}\\\\v=\sqrt{-2\times (-\mu g)\times d}\\\\v=\sqrt{2\mu gd}

Plug in the given values and find 'v'. This gives,

v=\sqrt{2\times 0.650\times 9.8\times 7.50}\\\\v=\sqrt{95.55}=9.77\ m/s

Now, using equation (1) and substituting the given values, we get:

0.014u=(0.014+0.110)\times 9.77\\\\u=\frac{0.124\times 9.77}{0.014}\\\\u=86.53\ m/s

Therefore, the speed of the clay immediately before impact is 86.53 m/s.

7 0
3 years ago
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