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PilotLPTM [1.2K]
3 years ago
5

Which statement accurately describes a similarity between Earth and Venus?

Physics
1 answer:
nikitadnepr [17]3 years ago
8 0

Your answer is B

There are many cases of volcanic eruptions on earth and it has been proven on Venus as well


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Definition of synthesis
Finger [1]
Combination or composition, in particular.
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3 years ago
A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
3 years ago
Most people can detect frequencies as high as 20 000 Hz. Assuming the speed of sound in air is 345 m/s, determine the wavelength
Gekata [30.6K]

Wavelength = speed / frequency

(345 m/s) / (20,000 Hz) = 0.017 m

6 0
3 years ago
Read 2 more answers
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
Newton's Law of Gravitation states that two bodies with masses m1 and m2 attract each other with a force F, where r is the dista
Marrrta [24]

Answer:

W = - 5.01 10¹⁰ J

Explanation:

Work is defined by the expression

      W = ∫ F.dr

Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product

      W = ∫ F dr

      W = G m₁m₂ ∫ 1 /r² dr

     W = G m₁ m₂2(-1 / r)

We evaluate between the lower limits r = Re and upper r = ∞

     W = G m₁m₂ (-1 / Re + 1 / ∞)

     W = - G m₁ m₂ / Re

Let's calculate

    W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶

    W = - 5.01 10¹⁰ J

4 0
3 years ago
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