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3 years ago

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A 5000-kg spacecraft is in a circular orbit 2000 km above the surface of mars. how much work must the spacecraft engines perform

to move the spacecraft to a circular orbit that is 4000 km above the surface?

1 answer:

Elden [556K]3 years ago

8 0

The spacecraft is orbiting at 2000 Km. The engine has to put it on 4000 km altitude, therefore the engine has to push it into another 2000 km, so that it will attain the 4000 km altitude.

Given. Gravitational pull of planet Mars g = 3.71 m/s²

Mass of the spacecraft m = 5000 kg

Formula: W = fd F = mg

W = mgh

W = (5000 Kg)(3. 71 m/s²)(2 x 10⁶m)

W = 3.71 x 10¹⁰ J

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storchak [24]

Bella’s average velocity is about 0.693 meters per second.

To find the average velocity, you must divide the distance by the change in time, which should look like v=d/t

Here is how you set up the equation-
v=6.1/8.8

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Hope this helped! If you have any questions about what I mentioned in my answer or explanation, feel free to comment on my answer and I’ll try to get back to you!

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IM SOO CONFUSED PLS HELP!! The mass of the nucleus is approximately EQUAL to the mass number multiplied by ____ Atomic Mass unit

nevsk [136]

Answer:

option a.

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We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.

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3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

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3 years ago

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Tems11 [23]

Answer:

temperature and mass

Explanation:

The higher the temperature of a given quantity of a substance, more is its thermal energy.

If a substance contains more mass, this also implies that the object has more particles in it . hence, it has high thermal energy.

<em><u>A</u></em><em><u>d</u></em><em><u>d</u></em><em><u>i</u></em><em><u>t</u></em><em><u>i</u></em><em><u>o</u></em><em><u>n</u></em><em><u>a</u></em><em><u>l</u></em><em><u> </u></em><em><u>I</u></em><em><u>n</u></em><em><u>f</u></em><em><u>o</u></em><em><u>r</u></em><em><u>m</u></em><em><u>a</u></em><em><u>t</u></em><em><u>i</u></em><em><u>o</u></em><em><u>n</u></em><em><u> </u></em>:

Temperature is a measure of the average kinetic energy of the particles of a substance.

The thermal energy of an object depends on three factors:

number of molecules in the object
temperature of the object.
thermal energy it has.

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3 years ago