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2 years ago

Earth is the _______ planet from the sun.

Physics

2 answers:

Alenkasestr [34]2 years ago

7 0

It is the third planet because it goes... Mercury, Venus, then Earth!!!

strojnjashka [21]2 years ago

6 0

Planet Earth is the third planet from the sun. The first four planets from the sun are Mercury, Venus, the Earth, and Mars, respectively.

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A 0.25-kg ball sits on the roof of a building that is 10 meters tall. Find the GPE. (Gravity = 9.8 on Earth)

Tasya [4]

Gravitational potential energy = mgh or mass times acceleration due to gravity times the height

Here the mass is 0.25kg, the height is 10m, and gravity is 9.8m/s^2 so...

GPE = (0.25)(10)(9.8)

GPE = 24.5 J

7 0

3 years ago

Does Ap Physics have anything to do with probablity?

padilas [110]

So I'm a junior. I am currently taking AP Calc BC and AP Physics B.

As of now, I'm not sure if I should take AP Probability and Statistics or Differential Equations/Calc III next year. Also, I'm debating between taking AP Physics C or AP Chemistry.

Which ones do you think would look better on a transcript? I heard that Diffeq/CalcIII is harder than AP ProbStat, but ProbStat is an AP course which will be weighted heavier. Also, should I take Physics C since i've taken Physics B this year already?

5 0

3 years ago

HELPPP PLEASEEE!!!!!

devlian [24]

Answer:

1) The speed of sound increases

2) 440 Hz

3) 29°C

4) 17°C

5) 434 Hz

6) 12 m/s

7) 17.3 m

Explanation:

1) The speed of sound increases

2) V = f×λ

f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz

3) V = f×λ

512 × 0.68 = 348.16 m/s

348.16 - 331 = 17.16

T = 17.16/0.6 = 28.6 ≈ 29°C

4) Increase in speed = 350 - 340 = 10

Increase in temperature = 10/0.6 = 16.67° ≈ 17°C

5) f = V/λ = 343/0.79 = 434 Hz

6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s

7) V = 331 + 0.6×25 = 346m/s

λ = 346/20 = 17.3 m

5 0

2 years ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

2 years ago

Read 2 more answers

Need help ??? Please

Yanka [14]

Is there information in the previous question which relates to this one?

8 0

2 years ago