Determine the distance between centers of two planets whose masses

Which of the following was NOT a department in Washington's first cabinet?

Bingel [31]

Answer:

I believe its C: Secretary of War. I hope this helped :)

Explanation:

6 0

3 years ago

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(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?

Mrrafil [7]

Answer:

69.69 g

Explanation:

Evaporation of water will take out latent heat of vaporization. Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg

Heat taken up by evaporating water

= 2260000 x m J

Heat lost by body

= mass x specific heat of body x drop in temperature

60 x 3500 x .750 ( specific heat of human body is 3.5 kJ/kg.k)

= 157500 J

Heat loss = heat gain

2260000 m= 157500

m = .06969 kg

= 69.69 g

5 0

4 years ago

Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying

agasfer [191]

Answer:

The charge density in the system is $4.25*10^4C/m$

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

$r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C$

We need to asume here the number of free electrons in a copper conductor, at which is generally of $8.5 *10^{28}m^{-3}$

The equation to find the current is

$I = VenA$

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

$I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)$

$I= 22.11a$

Now to find the linear charge density, we know that

$I = \frac{Q}{t} \rightarrow Q = It$

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

$V_d = \frac{l}{t} \rightarrow l = V_d t$

The charge density is defined as

$\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}$

Replacing our values

$\lambda = \frac{22.11}{5.20*10{-4}}$

$\lambda= 4.25*10^4C/m$

Therefore the charge density in the system is $4.25*10^4C/m$

5 0

4 years ago

A dogsled team has a total momentum of 1710 kg•m/s. If the team has a total mass of 450 kg, what is its speed?

HACTEHA [7]

Answer:

3.8 m/s

Explanation:

Momentum is the product of mass and velocity hence

Momentum=mv where m is mass and v is the velocity of an object

Making v the subject of the formula then

$v=\frac {momentum}{m}$

Substituting 1710 kgm/s for momentum and 450 Kg for mass we obtain

$v=\frac {1710 Kgm/s}{450 Kg}=3.8 m/s$

Hence the speed of dogsled team is 3.8 m/s

7 0

3 years ago

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Deimos completes one (circular) orbit of Mars in 1.26 days. The distance from Mars to Deimos is 2.35×107m. What is the centripet

Akimi4 [234]

Answer:

The centripetal acceleration of Deimos is $0.077m/s^{2}$ .

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity of Deimos and r is the orbital distance.

Notice that is necessary to determine the velocity first.

The speed of the Deimos can be found by means of the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$ (2)

Then, replacing Newton's second law in equation 2 it is gotten:

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (3)

However, a is the centripetal acceleration since Deimos almost describes a circular motion around Mars:

$a = \frac{v^{2}}{r}$ (4)

Replacing equation 4 in equation 3 it is gotten:

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (5)

Where v is the orbital speed, G is the gravitational constant, M is the mass of Mars, and r is the orbital radius.

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(6.39x10^{23}kg)}{2.35x10^{7}m}}$

$v = 1346m/s$

Finally, equation 4 can be used:

$a = \frac{(1346m/s)^{2}}{2.35x10^{7}m}$

$a = 0.077m/s^{2}$

Hence, the centripetal acceleration of Deimos is $0.077m/s^{2}$ .

7 0

3 years ago