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VikaD [51]
3 years ago
5

What is the number of Kelvins between the freezing point and the boiling point of water at a pressure of 1 atm?

Physics
1 answer:
Semmy [17]3 years ago
5 0

Answer:100 K

Explanation:

We know that Freezing Point of water is 0^{\circ}C at 1 atm

Converting it to Kelvin we get 273.15 K

Boiling Point of water is 100^{\circ}C

converting it to Kelvin we get 373.15 K

Difference in the number we get =373.15-273.15=100 K

Temperature difference is independent of degree and will remain same for Celsius and Kelvin

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A family took a trip in a car traveling East from Greensboro to Wilmington, NC. Use the Graph to answer the questions below.
kotykmax [81]

Answer:

1). Average speed = 1.5 m per second

2). Average velocity = 1.5 m per second

Explanation:

1). Since, speed is a scalar quantity

   Therefore, average speed of the trip = \frac{\text{Total distance covered}}{\text{Total time taken}}

    From the graph attached,

   Total distance covered = 10 + 10 + 20 + 0 + 20 + 30

                                           = 90 meters

   Total time taken = 60 seconds

    Average speed = \frac{90}{60}

                               = 1.5 meter per second

2). Velocity is a vector quantity.

    Therefore, average velocity = \frac{\triangle d}{\triangle t}

                                                   = \frac{d_{60}-d_0}{60-0}

                                                   = \frac{90-0}{60-0}

                                                   = 1.5 meter per second                        

7 0
3 years ago
A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck trave
Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s  = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

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3 years ago
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