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Sati [7]
3 years ago
7

If Mars were the same size as Mercury (instead of its actual size), which surface features would it have?

Physics
1 answer:
choli [55]3 years ago
6 0

Answer:

A surface that is full of craters, very similar to that one seen in mercury.

Explanation:

Mercury is the smaller planet in the solar system whit a very thin atmosphere. If Mars were the same size of mercury it will certainly have a surface full of craters.

When meteoroid (fragments of rock) reach the atmosphere of a planet, they get incinerate as a consequence of the friction between the object and the atmosphere. However, if the meteoroid has the necessary size it can reach the ground (at this point is known as meteorite).

In the case propose, it is most likely that bigger meteorite reach the surface of the planet, which leads to the result of a surface full of craters since the size of mars is not enough to maintain a very dense atmosphere due to the weak gravitational field.

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Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the
zlopas [31]

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

\nu = Frequency

\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

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3 years ago
How many milligrams are equivalent to 150 dekagrams?
dybincka [34]
 i believe  the answer is 1.5e+6

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4 0
3 years ago
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A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of
Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

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