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3 years ago

Please I need help on this

Physics

1 answer:

guapka [62]3 years ago

6 0

Answer: Generally, prevailing winds blow east-west rather than north-south. This happens because Earth's rotation generates what is known as the Coriolis effect. The Coriolis effect makes wind systems twist counter-clockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere.

Explanation: here u go

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Where in the ocean are u likely to find animals that make there own light

muminat

Usually in the deep sea and underwater caves where there is no light

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2 years ago

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If an area has all the wolves that it can support the wolf population has reached is what

Liono4ka [1.6K]

The wolf population in that area has reached its carrying capacity.

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3 years ago

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The mass of a meteoroid was found to be 1.24 × 107 kilograms. Write the mass in standard notation.

Musya8 [376]

Answer:

The value we are given in the question is 1.24 * 10^7. This form of writing number is called scientific notation. The standard notation is the normal, regular way of writing numbers. Scientific notation and standard notations are interchangeable.

1.24 * 10^7 written in standard notation will be = 1.24 * 10000000 = 12400000.

Thus, the mass of the meteoroid was 12400000 kg.

Explanation:

8 0

3 years ago

g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr

Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

$a = \frac{GM}{r^2}$

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

$\frac{da}{dr} = -2\frac{GM}{r^3}$

$da = -2\frac{GM}{r^3}dr$

At the same time since Newton's second law we know that:

$F_w = ma$

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

$F_W = m (\frac{GM}{r^2} ) = 600N$

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

$dF_W = mda$

$dF_W = m(-2\frac{GM}{r^3}dr)$

$dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})$

$dF_W = -2F_W(\frac{dr}{r})$

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

$dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})$

$dF_W = -0.3N$

Therefore there is a weight loss of 0.3N every kilometer.

4 0

3 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago