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aniked [119]
2 years ago
15

Thanks+ BRAINLIST only for correct answers

Physics
2 answers:
Maslowich2 years ago
8 0

Answer:

First) is b

Second) is true

Third) is C (not sure tho)

Explanation:

Because of their longer wavelengths, AM waves reflect off a layer of the upper atmosphere called the ionosphere.

antiseptic1488 [7]2 years ago
7 0
1st-B

2nd-A

3rd-C variation in frequency
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What does it mean for water to have a high heat capacity
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Its heat capacity is higher than that of any other liquid or solid, its specific heat being 1 cal / g, this means that to raise the temperature of 1 g of water by 1 ° C it is necessary to provide an amount of heat equal to a calorie . Therefore, the heat capacity of 1 g of water is equal to 1 cal / K.

Explanation:

The water has a very high heat capacity, a large amount of heat is necessary to raise its temperature 1.0 ° K. For biological systems this is very important because the cellular temperature is modified very little in response to metabolism. In the same way, aquatic organisms, if water did not possess that quality, would be very affected or would not exist.

This means that a body of water can absorb or release large amounts of heat, with little temperature change, which has a great influence on the weather (large bodies of water in the oceans take longer to heat and cool than the ground land). Its latent heats of vaporization and fusion (540 and 80 cal / g, respectively) are also exceptionally high.

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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
2 years ago
Read 2 more answers
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