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Basile [38]
3 years ago
5

20 point question B)

Physics
2 answers:
Lina20 [59]3 years ago
5 0
The correct answer is b
frez [133]3 years ago
4 0

Description:

As we know that the sun offers what most of the world has to go examples could be energy, air. As we also know that heating causes liquid and water that is frozen to evaporate to water vapour gas. Meaning the sun provides the energy necessary for evaporation.

Answer:  B

Please mark brainliest

<em><u>Hope this helps.</u></em>

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Explanation:

100 CM = 1 m

45 CM = 45 / 100 = 0.45 m

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A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a spe
Rasek [7]

Answer:

56.5\ \text{s}

21.13\ \text{m/s}

Explanation:

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

s = Displacement

Here the kinematic equations of motion are used

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}

Time the car is at constant velocity is 39 s

Time the car is decelerating is 5 s

Total time the car is in motion is 12.5+39+5=56.5\ \text{s}

Distance traveled

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}

s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2

s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}

The total displacement of the car is 156.25+975+62.5=1193.75\ \text{m}

Average velocity is given by

\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}

The average velocity of the car is 21.13\ \text{m/s}.

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a 20 of of ice at 0c is dropped into water at boiling point, specific heat capacity of water =4200 J/kg•c, sepesific latent heat
Contact [7]

Answer:

15 KJ

Explanation:

The quantity of heat (Q) required is given as:

Q = mcΔθ + mL

where m is the mass of ice, c is its specific heat capacity, L is its specific latent heat andΔθ is the change in temperature.

Given: m = 20g, temperature of ice = 0^{o} C, specific heat capacity of water = 4200 J/kg^{o} C, latent heat of fusion of ice = 3.3 x 10^(5) J/kg, temperature of water = 100^{o} C.

Q = m (cΔθ + L)

   = 0.02(4200 x (100) + 330000)

   = 0.02(420000 + 330000)

  = 0.02 (750000)

Q = 15000

Q = 15000 Joules

Q = 15KJ

The quantity of heat needed to complete the conversion is 15 KJ.

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