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3 years ago

How are earth and venus similar and different?

1 answer:

soldier1979 [14.2K]3 years ago

6 0

Similarities would be
They are both made of rock.
They are close in size.
They both have thick atmospheres.
They both have similar densities.
The strength of the gravity on their surfaces is similar.

Differences would be Venus and Earth are planets in our solar system, with Venus being the second closest planet and the Earth being the third closest to the sun. The mass of the earth is about 1.23 times the mass of Venus.

Being closer to the sun, Venus is a lot hotter than the Earth. While the average temperature on the earth is about 14 °C, that on Venus is over 460 °C. . Hope that helps

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Which is the graph of the polar equation r = 4 cos3 theta

Black_prince [1.1K]

Answer:

Im pretty sure it’s d

Explanation:

6 0

3 years ago

Read 2 more answers

Calculate the frequency of the wave shown below.

jenyasd209 [6]

$\color{skyblue}{ \underline{ \frak { \: option \: ( \: c \: ) = 2 \: hertz ✓}}}$

$\: \:$

Given :

Wavelength ( λ ) = 2 m

$\: \:$

Speed = 4 m/s

$\: \:$

We, have to find frequency :

$\:$

$\large \tt \: Frequency = \frac{Speed}{Wavelength ( \: λ \: )}$

$\: \:$

$\large \tt \: Frequency = \frac{4}{2}$

$\: \:$

$\large \tt \: Frequency = \cancel \frac{4}{2}$

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Hope Helps!

5 0

2 years ago

A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simp

masha68 [24]

To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as

$T = 2\pi \sqrt{\frac{l}{g}}$

Where

l = Length

g = Gravity

Re-arrange to find L,

$L = g (\frac{T}{2\pi})^2$

Our values are given as

$g = 9.81m/s$

$T = 10.1s$

Replacing,

$L = g (\frac{T}{2\pi})^2$

$L = (9.81) (\frac{10.1}{2\pi})^2$

$L = 25.348m$

Therefore the height would be 25.348m

5 0

3 years ago

What would happen if all the toilets were flushed at once?

Sliva [168]

Answer:

nothing would happen

Explanation:

5 0

4 years ago

Read 2 more answers

If The density of this stainless steel is7.85 g/cm3,specific heatis 0.5 J/g.K, melting pointis 1673K, heat of fusion s0.260J/kg.

kodGreya [7K]

Answer:

$\Delta H=687.4 J$

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

$\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J$

$\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J$

Therefore, the required heat is:

$\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J$

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

4 0

3 years ago